Date Редакция Категория comp Теги C

Уровень английского — элементарный, уровень вопросов — проверка на вшивость. Предназначены для устного решения, у меня же в некоторых местах возникли сомнения, которые я и проверил.

Директивы include и int в заголовке main() опущены составителями вопросов. Отсутствие у main() возвращаемого значения также оставлено без внимания. Кроме того, движок сайта иногда "съедает" символы формата и косую черту у '\n', так что бдите.

Здесь можно скачать вопросы без моих комментариев и исправлений.

Оригинальные задачи тестировались на компиляторе Turbo C/C++ под DOS (из этого, в частности следует, что в объяснениях принят size(int)==2). Я использую компилятор gcc (Ubuntu/Linaro 4.7.2-2ubuntu1) 4.7.2.

Каким будет результат выполнения следующего кода:

1.

main()
{
   int const *p=5;
   printf("%d",++(*p));
}

  • Answer: Compiler error: Cannot modify a constant value.
  • Explanation: p is a pointer to a "constant integer". But we tried to change the value of the "constant integer".

2.

main()
{
   char s[]="man";
   int i;
   for(i=0;s[i];i++)
   printf("\n%c%c%c%c",s[i],*(s+i),*(i+s),i[s]);
}

  • Answer:
    mmmm
    aaaa
    nnnn
  • Explanation: s[i], (i+s), (s+i), i[s] are all different ways of expressing the same idea. Generally array name is the base address for that array. Here s is the base address. i is the index number/displacement from the base address. So, indirecting it with * is same as s[i]. i[s] may be surprising. But in the case of C it is same as s[i].

3.

main()
{
   float me = 1.1;
   double you = 1.1;
   if(me==you)
       printf("I love U");
   else
       printf("I hate U");
}

  • Answer:
    I hate U
  • Explanation. For floating point numbers (float, double, long double) the values cannot be predicted exactly. Depending on the number of bytes, the precession with of the value represented varies. Float takes 4 bytes and long double takes 10 bytes. So float stores 0.9 with less precision than long double.
  • Rule of Thumb: Never compare or at-least be cautious when using floating point numbers with relational operators (== , >, <, <=, >=,!= ).

4.

main()
{
   static int var = 5;
   printf("%d ",var--);
   if(var)
       main();
}

  • Answer:
    5 4 3 2 1
  • Explanation. When static storage class is given, it is initialized once. The change in the value of a static variable is retained even between the function calls. Main is also treated like any other ordinary function, which can be called recursively.

5.

main()
{
   int c[ ]={2.8,3.4,4,6.7,5};
   int j,*p=c,*q=c;
   for(j=0;j<5;j++) {
      printf(" %d ",*c);
      ++q;     }
   for(j=0;j<5;j++){
      printf(" %d ",*p);
      ++p;     }
}

  • Answer:
    2 2 2 2 2 2 3 4 6 5
  • Explanation. Initially pointer c is assigned to both p and q. In the first loop, since only q is incremented and not c , the value 2 will be printed 5 times. In second loop p itself is incremented. So the values 2 3 4 6 5 will be printed.

6.

main()
{
   extern int i;
   i=20;
   printf("%d",i);
}

  • Answer: Linker Error : Undefined symbol '_i'
  • Explanation: extern storage class in the following declaration, extern int i; specifies to the compiler that the memory for i is allocated in some other program and that address will be given to the current program at the time of linking. But linker finds that no other variable of name i is available in any other program with memory space allocated for it. Hence a linker error has occurred .

7.

main()
{
   int i=-1,j=-1,k=0,l=2,m;
   m=i++ && j++ && k++ || l++;
   printf("%d %d %d %d %d",i,j,k,l,m);
}

  • Answer:
    0 0 1 3 1
  • Explanation. Logical operations always give a result of 1 or 0. And also the logical AND (&&) operator has higher priority over the logical OR (||) operator. So the expression ‘i++ && j++ && k++’ is executed first. The result of this expression is 0 (-1 && -1 && 0 = 0). Now the expression is 0 || 2 which evaluates to 1 (because OR operator always gives 1 except for ‘0 || 0’ combination- for which it gives 0). So the value of m is 1. The values of other variables are also incremented by 1.

8.

main()
{
   char *p;
   printf("%d %d",sizeof(*p),sizeof(p));
}

  • Answer:
    1 2
  • Explanation. The sizeof() operator gives the number of bytes taken by its operand. p is a character pointer, which needs one byte for storing its value (a character). Hence sizeof(*p) gives a value of 1. Since it needs two bytes to store the address of the character pointer sizeof(p) gives 2.
  • Комментарий. Результат зависит от компилятора. В моём случае sizeof(p) = 4

9.

main()
{
   int i=3;
   switch(i)
       {
       default:printf("zero");
       case 1: printf("one");
               break;
       case 2: printf("two");
               break;
       case 3: printf("three");
               break;
       } 
}

  • Answer:
    three
  • Explanation. The default case can be placed anywhere inside the loop. It is executed only when all other cases doesn't match.

10.

main()
{
    printf("%x",-1<<4);
}

  • Answer:
    fff0
  • Explanation. -1 is internally represented as all 1's. When left shifted four times the least significant 4 bits are filled with 0's.The %x format specifier specifies that the integer value be printed as a hexadecimal value.
  • Комментарий. У меня результат равен fffffff0, поскольку sizeof(int) = 4

11.

main()
{
   char string[]="Hello World";
   display(string);
}
void display(char *string)
{
   printf("%s",string);
}

  • Answer: Compiler Error : Type mismatch in redeclaration of function display
  • Explanation. In third line, when the function display is encountered, the compiler doesn't know anything about the function display. It assumes the arguments and return types to be integers, (which is the default type). When it sees the actual function display, the arguments and type contradicts with what it has assumed previously. Hence a compile time error occurs.
  • Комментарий. gcc ругается более конкретно: нет декларации «display» в этой области видимости.

12.

main()
{
   int c=- -2;
   printf("c=%d",c);
}

  • Answer: c=2;
  • Explanation. Here unary minus (or negation) operator is used twice. Same maths rules applies, ie. minus * minus= plus.
  • Note: However you cannot give like --2. Because -- operator can only be applied to variables as a decrement operator (eg., i--). 2 is a constant and not a variable.

13.

#define int char
main()
{
   int i=65;
   printf("sizeof(i)=%d",sizeof(i));
}

  • Answer: sizeof(i)=1
  • Explanation. Since the #define replaces the string int by the macro char

14.

main()
{
   int i=10;
   i=!i>14;
   printf ("i=%d",i);
}

  • Answer: i=0
  • Explanation. In the expression !i>14 , NOT (!) operator has more precedence than '>’ symbol. ! is a unary logical operator. !i (!10) is 0 (not of true is false). 0>14 is false (zero).

15.

main()
{
    char s[]={'a','b','c','\n','c','\0'};
    char *p,*str,*str1;
    p=&s[3];
    str=p;
    str1=s;
    printf("%d",++*p + ++*str1-32);
}

  • Answer: 77
  • Explanation: p is pointing to character '\n'. str1 is pointing to character 'a' ++p. "p is pointing to '\n' and that is incremented by one." the ASCII value of '\n' is 10, which is then incremented to 11. The value of ++p is 11. ++*str1, str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'. ASCII value of 'b' is 98. Now performing (11 + 98 – 32), we get 77 ("M").

16.

main()
{
    int a[2][2][2] = { {10,2,3,4}, {5,6,7,8} };
    int *p,*q;
    p=&a[2][2][2];
    *q=***a;
    printf("%d----%d",*p,*q);
}

  • Answer: SomeGarbageValue----10
  • Explanation: p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D(which you are not declared) it will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q, it will print first element of 3D array.
  • Комментарий. При компиляции выдаёт предупреждения: 3-я строка: отсутствуют фигурные скобки вокруг инициализатора [-Wmissing-braces]; 4-я строка: «q» is used uninitialized in this function [-Wuninitialized]. Из-за последнего, получаем во время выполнения Segmentation fault (core dumped). Исправить можно, честно направив указатель на первый элемент массива: q = &a[0][0][0];

17.

main()
{
struct xx
{
     int x=3;
     char name[]="hello";
};
struct xx *s;
printf("%d",s->x);
printf("%s",s->name);
}

  • Answer: Compiler Error
  • Explanation. You should not initialize variables in declaration

18.

int main()
{
   struct xx
   {
       int x;
       struct yy
       {
           char s;
           struct xx *p;
       };
       struct yy *q;
   };
}

  • Answer: Compiler Error
  • Explanation. The structure yy is nested within structure xx. Hence, the elements are of yy are to be accessed through the instance of structure xx, which needs an instance of yy to be known. If the instance is created after defining the structure, the compiler will not know about the instance relative to xx. Hence for nested structure yy you have to declare member.
  • Комментарий: компилятор, указывая на структуру yy, выдаёт предупреждение декларация ничего не описывает, которого легко избежать, записав *q непосредственно после определения структуры yy. Кроме того, выдаётся очевидное предупреждение: control reaches end of non-void function, т.к. целочисленная main() не возвращает никакого значения.

19.

main()
{
     printf("\nab");
     printf("\bsi");
     printf("\rha");
}

  • Answer: hai
  • Explanation: \n - newline, \b - backspace, \r - linefeed.
  • Комментарий: У меня вышло только ha. Причины: в особенностях работы \b и \r на разных системах. Бэкспейс не обязательно удаляет символ слева от курсора, он может лишь переводить курсор на одну позицию влево. В данном случае это не важно, результат один: asi. \r - возврат каретки - предполагает как простой перенос курсора в начало строки, что и предполагалось в объяснении, так и перенос с удалением всего содержимого строки, начиная с текущего положения курсора. Этот последний вариант у меня и реализован.

20.

main()
{
    int i=5;
    printf("%d%d%d%d%d",i++,i--,++i,--i,i);
}

  • Answer: 45545
  • Explanation. The arguments in a function call are pushed into the stack from left to right. The evaluation is by popping out from the stack, and the evaluation is from right to left, hence the result.
  • Комментарий. Объяснение правдоподобное, однако, у меня: 45555, поэтому заинтересовала такая трактовка: That arguments are taken from right to left and pre-increment/decrement are executed before any evaluation of the "printf(...)"; so first --i is executed... the ++i is executed... this makes the "i" keep the value of 5.

21.

#define square(x) x*x
main()
{
    int i;
    i = 64/square(4);
    printf("%d",i);
}

  • Answer: 64
  • Explanation: the macro call square(4) will substituted by 4*4 so the expression becomes i = 64/4*4. Since / and * has equal priority the expression will be evaluated as (64/4)*4 i.e. 16*4 = 64

22.

main()
{
    char *p="hai friends",*p1;
    p1=p;
    while(*p != '\0') ++*p++;
    printf("%s   %s",p,p1);
}

  • Answer: ibj!gsjfoet
  • Explanation: ++*p++ will be parse in the given order

    • *p that is value at the location currently pointed by p will be taken
    • ++*p the retrieved value will be incremented
    • when ; is encountered the location will be incremented that is p++ will be executed Hence, in the while loop initial value pointed by p is ‘h’, which is changed to ‘i’ by executing ++*p and pointer moves to point, ‘a’ which is similarly changed to ‘b’ and so on. Similarly blank space is converted to ‘!’. Thus, we obtain value in p becomes “ibj!gsjfoet” and since p reaches ‘\0’ and p1 points to p thus p1 does not print anything.
  • Комментарий. В результате получаем: Segmentation fault (core dumped), поскольку char *p="hai friends" -- строковый литерал, изменять который нельзя. Как только мы переделаем его в массив символов:

int main()
{
  char p[] = "hai friends";
  char *p0,*p1;
  p0 = p;
  p1 = p;
  while(*p0 !='\0') ++*p0++;
  printf("%s   %s",p0,p1);
  return 0;
}

то всё получится. Кроме того, выводится содержимое p1, т. к. p указывает на конец строки.

23.

#define a 10
main()
{
    #define a 50
    printf("%d",a);
}

  • Answer: 50
  • Explanation. The preprocessor directives can be redefined anywhere in the program. So the most recently assigned value will be taken.

24.

#define clrscr() 100
main()
{
    clrscr();
    printf("%d\n",clrscr());
}

  • Answer: 100
  • Explanation. Preprocessor executes as a separate pass before the execution of the compiler. So textual replacement of clrscr() to 100 occurs.The input program to compiler looks like this:
main()
{
    100;
    printf("%d\n",100);
}
  • Note: 100; is an executable statement but with no action. So it doesn't give any problem.

25.

main()
{
  printf("%p",main);
}

  • Answer: Some address will be printed.
  • Explanation. Function names are just addresses (just like array names are addresses). main() is also a function. So the address of function main will be printed. %p in printf specifies that the argument is an address. They are printed as hexadecimal numbers.

26.

main()
{
    clrscr();
}
clrscr();

  • Answer: No output/error
  • Explanation. The first clrscr() occurs inside a function. So it becomes a function call. In the second clrscr(); is a function declaration (because it is not inside any function).
  • Комментарий. Конкретно, на первый clrscr() компилятор ругается: нет декларации «clrscr» в этой области видимости. На второй: expected constructor, destructor, or type conversion before «;» token.

27.

enum colors {BLACK,BLUE,GREEN};
main()
{
    printf("%d..%d..%d",BLACK,BLUE,GREEN);
    return(1);
}

  • Answer: 0..1..2
  • Explanation: enum assigns numbers starting from 0, if not explicitly defined.

28.

main()
{
   char far *farther, *farthest;

   printf("%d..%d",sizeof(farther),sizeof(farthest));
}

  • Answer: 4..2
  • Explanation: the second pointer is of char type and not a far pointer.
  • Комментарий. gcc не поддерживает far-указатели и, как следствие, выдает ошибки. Все указатели для gcc имеют размер 4.

29.

main()
{
    int i=400, j=300;
    printf("%d..%d");
}

  • Answer: 400..300
  • Explanation: printf takes the values of the first two assignments of the program. Any number of printf's may be given. All of them take only the first two values. If more number of assignments given in the program, then printf will take garbage values.
  • У меня: gcc предупреждает:
    • format «%d» expects a matching «int» argument [-Wformat]
    • неиспользуемая переменная «i» [-Wunused-variable]
    • неиспользуемая переменная «j» [-Wunused-variable] и печатает каждый раз разный мусор.

30.

main()
{
    char *p;
    p="Hello";
    printf("%c\n",*&*p);
}

  • Answer: H
  • Explanation: * is a dereference operator & is a reference operator. They can be applied any number of times provided it is meaningful. Here p points to the first character in the string "Hello". *p dereferences it and so its value is H. Again & references it to an address and * dereferences it to the value H.

31.

main()
{
    int i=1;
    while (i<=5)
    {
        printf("%d",i);
        if (i>2)
             goto here;
        i++;
    }
}
fun()
{
   here:
   printf("PP");
}

  • Answer: Compiler error: Undefined label 'here' in function main
  • Explanation. Labels have functions scope, in other words The scope of the labels is limited to functions. The label 'here' is available in function fun() Hence it is not visible in function main.

32.

main()
{
    static char names[5][20]={"pascal", "ada", "cobol", "fortran", "perl"};
    int i;
    char *t;
    t=names[3];
    names[3]=names[4];
    names[4]=t;
    for (i=0;i<=4;i++)
         printf("%s",names[i]);
}

  • Answer: Compiler error: Lvalue required in function main
  • Explanation. Array names are pointer constants. So it cannot be modified.
  • У меня: в строке names[3]=names[4]; и следующей ошибка вида: incompatible types when assigning to type «char[20]» from type «char *»

33.

main()
{
    int i=5;
    printf("%d",i++ + ++i);
}

  • Answer: Output Cannot be predicted exactly.
  • Explanation: Side effects are involved in the evaluation of i.
  • У меня: предупреждение: операция над «i» может дать неопределенный результат [-Wsequence-point] и результат: 12. Видимо, аргумент printf оценивается справа налево. Само i после преобразований равно 7.

34.

main()
{
    int i=5;
    printf("%d",i+++++i);
}

  • Answer: Compiler Error
  • Explanation. The expression i+++++i is parsed as i ++ ++ + i which is an illegal combination of operators.
  • У меня: в строке с printf: ошибка: lvalue required as increment operand

35.

main()
{
    int i=1,j=2;
    switch(i)
    {
     case 1:  printf("GOOD");
              break;
     case j:  printf("BAD");
              break;
    }
}

  • Answer: Compiler Error: Constant expression required in function main.
  • Explanation. The case statement can have only constant expressions (this implies that we cannot use variable names directly so an error).
  • Note: Enumerated types can be used in case statements.

36.

main()
{
    int i;
    printf("%d",scanf("%d",&i));  // value 10 is given as input here
}

  • Answer: 1
  • Explanation: Scanf returns number of items successfully read and not 1/0. Here 10 is given as input which should have been scanned successfully. So number of items read is 1.

37.

#define f(g,g2) g##g2
main()
{
    int var12=100;
    printf("%d",f(var,12));
}

  • Answer: 100

38.

main()
{
    int i=0;

    for( ; i++; printf("%d",i));
    printf("%d",i);
}

  • Answer: 1
  • Explanation: before entering into the for loop the checking condition is "evaluated". Here it evaluates to 0 (false) and comes out of the loop, and i is incremented (note the semicolon after the for loop).

39.

main()
{
    char s[]={'a','b','c','\n','c','\0'};
    char *p,*str,*str1;
    p=&s[3];
    str=p;
    str1=s;
    printf("%d",++*p + ++*str1-32);
}

  • Answer: M
  • Explanation: p is pointing to character '\n'. str1 is pointing to character 'a', ++p: "p is pointing to '\n' and that is incremented by one". The ASCII value of '\n' is 10, then it is incremented to 11. The value of ++p is 11. ++*str1: "str1 is pointing to 'a' that is incremented by 1 and it becomes 'b'". ASCII value of 'b' is 98. Both 11 and 98 is added and result is subtracted from 32. I.e. (11+98-32)=77("M").

40.

main()
{
    struct xx
    {
         int x=3;
         char name[]="hello";
    };
    struct xx *s=malloc(sizeof(struct xx));
    printf("%d",s->x);
    printf("%s",s->name);
}

  • Answer: Compiler Error
  • Explanation. Initialization should not be done for structure members inside the structure declaration

41.

main()
{
    extern int i;
    i=20;
    printf("%d",sizeof(i));
}

  • Answer: Linker error: undefined symbol '_i'.
  • Explanation: extern declaration specifies that the variable i is defined somewhere else. The compiler passes the external variable to be resolved by the linker. So compiler doesn't find an error. During linking the linker searches for the definition of i. Since it is not found the linker flags an error.
  • Комментарий: та же ошибка называется: undefined reference to 'i'.

42.

main()
{
    printf("%d", out);
}

int out=100;

  • Answer: Compiler error: undefined symbol out in function main.
  • Explanation. The rule is that a variable is available for use from the point of declaration. Even though a is a global variable, it is not available for main. Hence an error.

43.

main()
{
  extern out;
  printf("%d", out);
}
  int out=100;

  • Answer: 100
  • Explanation: This is the correct way of writing the previous program.

44.

main()
{
  show();
}
void show()
{
  printf("I'm the greatest");
}

  • Answer: Compiler error: Type mismatch in redeclaration of show.
  • Explanation. When the compiler sees the function show it doesn't know anything about it. So the default return type (i.e., int) is assumed. But when compiler sees the actual definition of show mismatch occurs since it is declared as void. Hence the error. The solutions are as follows:
    • declare void show() in main().
    • define show() before main().
    • declare extern void show() before the use of show().
  • Комментарий. Существенные предупреждения:
    • неявная декларация функции «show»
    • несовместимые типы для «show»

45.

main()
{
  int a[2][3][2] = {{{2,4},{7,8},{3,4}},{{2,2},{2,3},{3,4}}};
  printf("%u %u %u %d \n",a,*a,**a,***a);
  printf("%u %u %u %d \n",a+1,*a+1,**a+1,***a+1);
}

  • Answer:
    100, 100, 100, 2
    114, 104, 102, 3
  • Explanation. The given array is a 3-D one. It can also be viewed as a 1-D array (см. ниже).
    Thus, for the first printf statement a, a, a give address of first element, since the indirection a gives the value. Hence, the first line of the output.
    For the second printf a+1 increases in the third dimension thus points to value at 114,
    a+1 increments in second dimension thus points to 104,
    a+1 increments the first dimension thus points to 102 and
    a+1 first gets the value at first location and then increments it by 1. Hence, the output.
  • Комментарий. Лучше использовать для указателей спецификаторы %p. Размерности я считаю наоборот - от старшей к младшей. Так или иначе, но a+1 имеет адрес 112, а не 114.
2 4 7 8 3 4 2 2 2 3 3 4
100 102 104 106 108 110 112 114 116 118 120 122

46.

main()
{
  int a[] = {10,20,30,40,50},j,*p;
  for(j=0; j<5; j++) {
    printf("%d",*a); 
    a++;
  }
  p = a;
  for(j=0; j<5; j++) {
    printf("%d" ,*p); 
    p++;
  }
}

  • Answer: Compiler error: lvalue required.
  • Explanation. Error is in line with statement a++. The operand must be an lvalue and may be of any of scalar type for the any operator, array name only when subscripted is an lvalue. Simply array name is a non-modifiable lvalue.

47.

main()
{
  static int a[] = {0,1,2,3,4};
  int *p[] = {a,a+1,a+2,a+3,a+4};
  int **ptr =  p;
  ptr++;
  printf("\n %d  %d  %d", ptr-p, *ptr-a, **ptr); 
  *ptr++;
  printf("\n %d  %d  %d", ptr-p, *ptr-a, **ptr); 
  *++ptr;
  printf("\n %d  %d  %d", ptr-p, *ptr-a, **ptr); 
  ++*ptr;
  printf("\n %d  %d  %d", ptr-p, *ptr-a, **ptr); 
}

  • Answer:
    111
    222
    333
    344
  • Explanation. Let us consider the array and the two pointers with some address
a 0 1 2 3 4
100 102 104 106
p 100 102 104 106 108
1000 1002 1004 1006
ptr 1000

After execution of the instruction ptr++ value in ptr becomes 1002, if scaling factor for integer is 2 bytes. Now ptr – p is value in ptr – starting location of array p, (1002 – 1000) / (scaling factor) = 1, ptr – a = value at address pointed by ptr – starting value of array a, 1002 has a value 102 so the value is (102 – 100)/(scaling factor) = 1, *ptr is the value stored in the location pointed by the pointer of ptr = value pointed by value pointed by 1002 = value pointed by 102 = 1. Hence the output of the first printf is 1, 1, 1.

After execution of ptr++ increments value of the value in ptr by scaling factor, so it becomes 1004. Hence, the outputs for the second printf are ptr – p = 2, ptr – a = 2, **ptr = 2.

After execution of ++ptr increments value of the value in ptr by scaling factor, so it becomes 1006. Hence, the outputs for the third printf are ptr – p = 3, ptr – a = 3, **ptr = 3.

After execution of ++ptr value in ptr remains the same, the value pointed by the value is incremented by the scaling factor. So the value in array p at location 1006 changes from 106 to 108,. Hence, the outputs for the fourth printf are ptr – p = 1006 – 1000 = 3, ptr – a = 108 – 100 = 4, **ptr = 4.

48.

main( )
{
  char  *q;
  int  j;
  for (j=0; j<3; j++) scanf("%s" ,(q+j));
  for (j=0; j<3; j++) printf("%c" ,*(q+j));
  for (j=0; j<3; j++) printf("%s" ,(q+j));
}

  • Explanation. Here we have only one pointer to type char and since we take input in the same pointer thus we keep writing over in the same location, each time shifting the pointer value by 1.
    Suppose the inputs are MOUSE, TRACK and VIRTUAL. Then for the first input suppose the pointer starts at location 100 then the input one is stored as
    M O U S E \0
    When the second input is given the pointer is incremented as j value becomes 1, so the input is filled in memory starting from 101.
    M T R A C K \0
    The third input starts filling from the location 102
    M T V I R T U A L \0
    This is the final value stored.
    The first printf prints the values at the position q, q+1 and q+2 = M T V
    The second printf prints three strings starting from locations q, q+1, q+2 i.e MTVIRTUAL, TVIRTUAL and VIRTUAL.

49.

main()
{
  void *vp;
  char ch = 'g', *cp = "goofy";
  int j = 20;
  vp = &ch;
  printf("%c", *(char *)vp);
  vp = &j;
  printf("%d",*(int *)vp);
  vp = cp;
  printf("%s",(char *)vp + 3);
}

  • Answer: g20fy
  • Explanation. Since a void pointer is used it can be type casted to any other type pointer. vp = &ch stores address of char ch and the next statement prints the value stored in vp after type casting it to the proper data type pointer. the output is ‘g’. Similarly, the output from second printf is ‘20’. The third printf statement type casts it to print the string from the 4th value hence the output is ‘fy’.

50.

main ()
{
  static char *s[]  = {"black", "white", "yellow", "violet"};
  char **ptr[] = {s+3, s+2, s+1, s}, ***p;
  p = ptr;
  **++p;
  printf("%s",*--*++p + 3);
}

  • Answer: ck
  • Explanation. In this problem we have an array of char pointers pointing to start of 4 strings. Then we have ptr which is a pointer to a pointer of type char and a variable p which is a pointer to a pointer to a pointer of type char. p hold the initial value of ptr, i.e. p = s+3. The next statement increment value in p by 1, thus now value of p = s+2. In the printf statement the expression is evaluated *++p causes gets value s+1 then the pre decrement is executed and we get s+1 – 1 = s . the indirection operator now gets the value from the array of s and adds 3 to the starting address. The string is printed starting from this position. Thus, the output is ‘ck’.

51.

main()
{
  int  i, n;
  char *x = "girl";
  n = strlen(x);
  *x = x[n];
  for(i=0; i<n; ++i)
  {
    printf("%s\n",x);
    x++;
  }
}

  • Answer:
    (blank space)
    irl
    rl
    l
  • Explanation. Here a string (a pointer to char) is initialized with a value "girl". The strlen function returns the length of the string, thus n has a value 4. The next statement assigns value at the nth location (‘\0’) to the first location. Now the string becomes "\0irl" . Now the printf statement prints the string after each iteration it increments it starting position. Loop starts from 0 to 4. The first time x[0] = ‘\0’ hence it prints nothing and pointer value is incremented. The second time it prints from x[1] i.e "irl" and the third time it prints "rl" and the last time it prints "l" and the loop terminates.

52.

int i,j;
for(i=0;i<=10;i++)
{
  j+=5;
  assert(i<5);
}

  • Answer:
    Runtime error: Abnormal program termination.
    assert failed (i<5), <file name>,<line number>
  • Explanation: asserts are used during debugging to make sure that certain conditions are satisfied. If assertion fails, the program will terminate reporting the same. After debugging use, #undef NDEBUG, and this will disable all the assertions from the source code. Assertion is a good debugging tool to make use of.

53.

main()
{
  int i=-1;
  +i;
  printf("i = %d, +i = %d \n",i,+i);
}

  • Answer: i = -1, +i = -1
  • Explanation. Unary + is the only dummy operator in C. Where-ever it comes you can just ignore it just because it has no effect in the expressions (hence the name dummy operator).

54.

int main()
{
  main();
}

  • Answer: Runtime error : Stack overflow.
  • Explanation: main function calls itself again and again. Each time the function is called its return address is stored in the call stack. Since there is no condition to terminate the function call, the call stack overflows at runtime. So it terminates the program and results in an error.
  • Комментарий. Объяснение звучит логично, но у меня Program received signal SIGSEGV, Segmentation fault.

55.

main()
{
  char *cptr,c;
  void *vptr,v;
  c=10; v=0;
  cptr=&c; vptr=&v;
  printf("%c%v",c,v);
}

  • Answer: Compiler error (at line number 4): size of v is Unknown.
  • Explanation. You can create a variable of type void * but not of type void, since void is an empty type. In the second line you are creating variable vptr of type void * and v of type void hence an error.

56.

main()
{
  char *str1="abcd";
  char str2[]="abcd";
  printf("%d %d %d",sizeof(str1),sizeof(str2),sizeof("abcd"));
}

  • Answer: 2 5 5
  • Explanation. In first sizeof, str1 is a character pointer so it gives you the size of the pointer variable. In second sizeof the name str2 indicates the name of the array whose size is 5 (including the '\0' termination character). The third sizeof is similar to the second one.

57.

main()
{
  char not;
  not=!2;
  printf("%d",not);
}

  • Answer: 0
  • Explanation. ! is a logical operator. In C the value 0 is considered to be the boolean value FALSE, and any non-zero value is considered to be the boolean value TRUE. Here 2 is a non-zero value so TRUE. !TRUE is FALSE (0) so it prints 0.

58.

#define FALSE -1
#define TRUE   1
#define NULL   0
main()
{
  if(NULL)
    puts("NULL");
  else if(FALSE)
    puts("TRUE");
  else
    puts("FALSE");
}

  • Answer: TRUE
  • Explanation. The input program to the compiler after processing by the preprocessor is,
main()
{
  if(0)
    puts("NULL");
  else if(-1)
    puts("TRUE");
  else
    puts("FALSE");
}

Preprocessor doesn't replace the values given inside the double quotes. The check by if condition is boolean value false so it goes to else. In second if -1 is boolean value true hence "TRUE" is printed.

59.

main()
{
  int k=1;
  printf("%d==1 is ""%s",k,k==1?"TRUE":"FALSE");
}

  • Answer: 1==1 is TRUE
  • Explanation. When two strings are placed together (or separated by white-space) they are concatenated (this is called as "stringization" operation). So the string is as if it is given as "%d==1 is %s". The conditional operator( ?: ) evaluates to "TRUE".

60.

main()
{
  int y;
  scanf("%d",&y); // input given is 2000
  if( (y%4==0 && y%100 != 0) || y%100 == 0 )
     printf("%d is a leap year");
  else
     printf("%d is not a leap year");
}

  • Answer: 2000 is a leap year

61.

#define max 5
#define int arr1[max]
main()
{
  typedef char arr2[max];
  arr1 list={0,1,2,3,4};
  arr2 name="name";
  printf("%d %s",list[0],name);
}

  • Answer: Compiler error (in the line arr1 list = {0,1,2,3,4})
  • Explanation: arr2 is declared of type array of size 5 of characters. So it can be used to declare the variable name of the type arr2. But it is not the case of arr1. Hence an error.
  • Rule of Thumb: #defines are used for textual replacement whereas typedefs are used for declaring new types.

62.

int i=10;
main()
{
  extern int i;
  {
    int i=20;
    {
      const volatile unsigned i=30;
      printf("%d",i);
    }
    printf("%d",i);
  }
  printf("%d",i);
}

  • Answer: 30,20,10
  • Explanation: '{' introduces new block and thus new scope. In the innermost block i is declared as, const volatile unsigned which is a valid declaration. i is assumed of type int. So printf prints 30. In the next block, i has value 20 and so printf prints 20. In the outermost block, i is declared as extern, so no storage space is allocated for it. After compilation is over the linker resolves it to global variable i (since it is the only variable visible there). So it prints i's value as 10.

63.

main()
{
  int *j;
  {
    int i=10;
    j=&i;
  }
  printf("%d",*j);
}

  • Answer: 10
  • Explanation. The variable i is a block level variable and the visibility is inside that block only. But the lifetime of i is lifetime of the function so it lives upto the exit of main function. Since the i is still allocated space, *j prints the value stored in i since j points i.

64.

main()
{
  int i=-1;
  -i;
  printf("i = %d, -i = %d \n",i,-i);
}

  • Answer: i = -1, -i = 1
  • Explanation: -i is executed and this execution doesn't affect the value of i. In printf first you just print the value of i. After that the value of the expression -i = -(-1) is printed.

65.

main()
{
  const int i=4;
  float j;
  j = ++i;
  printf("%d  %f", i,++j);
}

  • Answer: Compiler error
  • Explanation: i is a constant. You cannot change the value of constant.

66.

main()
{
  int a[2][2][2] = { {10,2,3,4}, {5,6,7,8}  };
  int *p,*q;
  p=&a[2][2][2];
  *q=***a;
  printf("%d..%d",*p,*q);
}

  • Answer: garbagevalue..10
  • Explanation: p=&a[2][2][2] you declare only two 2D arrays, but you are trying to access the third 2D (which you are not declared). It will print garbage values. *q=***a starting address of a is assigned integer pointer. Now q is pointing to starting address of a. If you print *q it will print first element of 3D array.

67.

main()
{
  register i=5;
  char j[]= "hello";
  printf("%s  %d",j,i);
}

  • Answer: hello 5
  • Explanation: if you declare i as register compiler will treat it as ordinary integer and it will take integer value. i value may be stored either in register or in memory.

68.

main()
{
  int i=5,j=6,z;
  printf("%d",i+++j);
}

  • Answer: 11
  • Explanation: the expression i+++j is treated as (i++ + j)

69.

struct aaa{
  struct aaa *prev;
  int i;
  struct aaa *next;
};

main()
{
  struct aaa abc,def,ghi,jkl;
  int x=100;
  abc.i=0;abc.prev=&jkl;
  abc.next=&def;
  def.i=1;def.prev=&abc;def.next=&ghi;
  ghi.i=2;ghi.prev=&def;
  ghi.next=&jkl;
  jkl.i=3;jkl.prev=&ghi;jkl.next=&abc;
  x=abc.next->next->prev->next->i;
  printf("%d",x);
}

  • Answer: 2
  • Explanation: above all statements form a double circular linked list; abc.next->next->prev->next->i — this one points to "ghi" node the value of at particular node is 2.

70.

struct point
{
  int x;
  int y;
};

struct point origin,*pp;

main()
{
  pp=&origin;
  printf("origin is(%d%d)\n",(*pp).x,(*pp).y);
  printf("origin is (%d%d)\n",pp->x,pp->y);
} 

  • Answer:
    origin is(0,0)
    origin is(0,0)
  • Explanation. Since structure point is globally declared x & y are initialized as zeroes. pp is a pointer to structure, so we can access the elements of the structure either with arrow mark or with indirection operator.

71.

main()
{
  int i=_l_abc(10);
  printf("%d\n",--i);
}

int _l_abc(int i)
{
 return(i++);
}

  • Answer: 9
  • Explanation: return(i++) it will first return i and then increments, i.e. 10 will be returned.

72.

main()
{
  char *p;
  int *q;
  long *r;
  p=q=r=0;
  p++;
  q++;
  r++;
  printf("%p...%p...%p",p,q,r);
}

  • Answer: 0001...0002...0004
  • Explanation: increment operator when applied to pointers increments address according to their corresponding data-types.
  • Комментарий: компилятор выдает ошибки в 6-й строке:
    • ошибка: некорректное преобразование из «long int*» в «int*» [-fpermissive]
    • ошибка: cannot convert «int*» to «char*» in assignment Если не приравнивать все указатели скопом (записать, например, p=0; q=0; r=0;), то получим ожидаемое: 0x1...0x4...0x4

73.

main()
{
  char c=' ',x,convert(z);
  getc(c);
  if ((c>='a') && (c<='z'))
    x=convert(c);
  printf("%c",x);
}

convert(z)
{
  return z-32;
}

  • Answer: Compiler error
  • Explanation: declaration of convert and format of getc() are wrong.
  • Комментарий: А ещё "нет декларации «z» в этой области видимости" в 3-й строке.

74.

main(int argc, char **argv)
{
  printf("enter the character");
  getchar();
  sum(argv[1],argv[2]);
}

sum(num1,num2)
int num1,num2;
{
  return num1+num2;
}

  • Answer: Compiler error.
  • Explanation. argv[1] & argv[2] are strings. They are passed to the function sum without converting it to integer values.
  • Комментарий. А ещё у sum() не задан перед main() прототип и не указаны типы данных аргументов (а то, как указано, неверно).

75.

int one_d[]={1,2,3};
main()
{
  int *ptr; 
  ptr=one_d;
  ptr+=3;
  printf("%d",*ptr);
}

  • Answer: garbage value
  • Explanation: ptr pointer is pointing to out of the array range of one_d.

76.

aaa() {
  printf("hi");
}
bbb() {
  printf("hello");
}
ccc() {
  printf("bye");
}

main()
{
  int (*ptr[3])();
  ptr[0]=aaa;
  ptr[1]=bbb;
  ptr[2]=ccc;
  ptr[2]();
}

  • Answer: bye
  • Explanation. ptr is array of pointers to functions of return type int.ptr[0] is assigned to address of the function aaa. Similarly ptr[1] and ptr[2] for bbb and ccc respectively. ptr2 is in effect of writing ccc(), since ptr[2] points to ccc.

77.

main()
{
  FILE *ptr;
  char i;
  ptr=fopen("zzz.c","r");
  while((i=fgetc(ptr))!=EOF)
    printf("%c",i);
}

  • Answer: contents of zzz.c followed by an infinite loop
  • Explanation. The condition is checked against EOF, it should be checked against NULL.
  • Комментарий. Не согласен с обоими:) Будет выведено содержимое zzz.c без каких-либо бесконечных циклов и проверка на не-EOF вполне логична.

78.

main()
{
  int i =0;j=0;
  if(i && j++)
    printf("%d..%d",i++,j);
  printf("%d..%d,i,j);
}

  • Answer: 0..0
  • Explanation. The value of i is 0. Since this information is enough to determine the truth value of the boolean expression. So the statement following the if statement is not executed. The values of i and j remain unchanged and get printed.

79.

main()
{
  int i;
  i = abc();
  printf("%d",i);
}

abc()
{
  _AX = 1000;
}

  • Answer: 1000
  • Explanation. Normally the return value from the function is through the information from the accumulator. Here _AX is the pseudo global variable denoting the accumulator. Hence, the value of the accumulator is set 1000 so the function returns value 1000.
  • Комментарий. Программа и объяснение её работы выглядят горячечным бредом. _AX в функции не определена. Можно попытаться исправить код, например, так (результат: 1000)
#include <stdio.h>

int abc()
{
  register int _AX = 1000;
  return _AX;
}

int main()
{
  int i;
  i = abc();
  printf("%d",i);
  return 0;
}

но получается трюизм.

80.

int i;
main()
{
  int t;
  for (t=4; scanf("%d",&i)-t; printf("%d\n",i))
    printf("%d--",t--);
}
// If the inputs are 0,1,2,3 find the o/p

  • Answer:
    4--0
    3--1
    2--2
  • Explanation. Let us assume some x=scanf("%d",&i)-t the values during execution will be, t i x
    4 0 -4
    3 1 -2
    2 2 0

81.

main()
{
  int a=0; int b=20; char x=1; char y=10;
  if(a,b,x,y)
    printf("hello");
}

  • Answer: hello
  • Explanation. The comma operator has associativity from left to right. Only the rightmost value is returned and the other values are evaluated and ignored. Thus the value of last variable y is returned to check in if. Since it is a non zero value if becomes true so, "hello" will be printed.
  • Комментарий. gcc честно выдаст предупреждения о том, что три из операндов операции "," не будут иметь эффекта.

82.

main()
{
  unsigned int i;
  for(i=1; i>-2; i--)
    printf("c aptitude");
}

  • Answer:
  • Explanation: i is an unsigned integer. It is compared with a signed value. Since the both types doesn't match, signed is promoted to unsigned value. The unsigned equivalent of -2 is a huge value so condition becomes false and control comes out of the loop.

83.

main()
{
  char *p;
  p="%d\n";
  p++;
  p++;
  printf(p-2,300);
}

  • Answer: 300
  • Explanation. The pointer points to % since it is incremented twice and again decremented by 2, it points to '%d\n' and 300 is printed.
  • Комментарий. С поправкой на ругань компилятора по поводу приравнивания строки и указателя на символ.

84.

main()
{
  char a[100];
  a[0]='a';a[1]='b';a[2]='c';a[4]='d';
  abc(a);
}

abc(char a[])
{
  a++; 
  printf("%c",*a);
  a++;
  printf("%c",*a);
}

  • Answer: bc
  • Explanation. The base address is modified only in function and as a result a points to 'b' then after incrementing to 'c' so 'bc' will be printed.

85.

func(a,b)
int a,b;
{
  return(a=(a==b));
}

main()
{
  int process(),func();
  printf("The value of process is %d !\n ",process(func,3,6));
}

process(pf,val1,val2)
int (*pf) ();
int val1,val2;
{
  return((*pf) (val1,val2));
}

  • Answer: The value if process is 0 !
  • Explanation. The function 'process' has 3 parameters - 1, a pointer to another function 2 and 3, integers. When this function is invoked from main, the following substitutions for formal parameters take place: func for pf, 3 for val1 and 6 for val2. This function returns the result of the operation performed by the function 'func'. The function func has two integer parameters. The formal parameters are substituted as 3 for a and 6 for b. since 3 is not equal to 6, a==b returns 0. therefore the function returns 0 which in turn is returned by the function 'process'.
  • Комментарий. Вместо оформления в стиле K&R хотелось бы видеть более стандартное
#include <stdio.h>

int main()
{
    int process(),func();
    printf("The value of process is %d !\n ",process(func,3,6));
    return 0;
}

int func(int a,int b)
{ return( a= (a==b) ); }

int process(int (*pf)(), int val1, int val2)
{ return((*pf) (val1,val2)); }

86.

void main()
{
  static int i=5;
  if(--i) {
    main();
    printf("%d ",i);
  }
}

  • Answer: 0 0 0 0
  • Explanation. The variable "i" is declared as static, hence memory for i will be allocated for only once, as it encounters the statement. The function main() will be called recursively unless i becomes equal to 0, and since main() is recursively called, so the value of static i i.e., 0 will be printed every time the control is returned.

87.

void main()
{
  int k=ret(sizeof(float));
  printf("\n here value is %d",++k);
}

int ret(int ret)
{
  ret += 2.5;
  return(ret);
}

  • Answer: Here value is 7
  • Explanation. Firstly, the function ret() is called in which the sizeof(float) ie., 4 is passed, after the first expression the value in ret will be 6, as ret is integer hence the value stored in ret will have implicit type conversion from float to int. The ret is returned in main() it is printed after and preincrement. The int ret(int ret), i.e., the function name and the argument name can be the same.

88.

main()
{
  char a[]="12345\0";
  int i=strlen(a);
  printf("here in 3 %d\n",++i);
}

  • Answer: here in 3 6
  • Explanation. The char array 'a' will hold the initialized string, whose length will be counted from 0 till the null character. Hence the 'i' will hold the value equal to 5, after the pre-increment in the printf statement, the 6 will be printed.

89.

main()
{
  unsigned giveit=-1;
  int gotit;
  printf("%u ",++giveit);
  printf("%u \n",gotit=--giveit);
}

  • Answer: 0 65535
  • Комментарий. Приведенные значения вывода соответствуют 2-байтному int.

90.

main()
{
  int i;
  char a[]="\0";
  if (printf("%s\n",a))
    printf("Ok here \n");
  else
    printf("Forget it\n");
}

  • Answer: Ok here
  • Explanation. printf will return how many characters does it print. Hence printing a null character returns 1 which makes the if statement true, thus "Ok here" is printed.

91.

main()
{
  void *v;
  int integer=2;
  int *i=&integer;
  v=i;
  printf("%d",(int*)*v);
}

  • Answer: Compiler Error. We cannot apply indirection on type void.
  • Explanation. Void pointer is a generic pointer type. No pointer arithmetic can be done on it. Void pointers are normally used for,
    • Passing generic pointers to functions and returning such pointers.
    • As a intermediate pointer type.
    • Used when the exact pointer type will be known at a later point of time.
  • Комментарий. Ошибка заключается в разыменовании обобщенного указателя в printf(). Её можно исправить, изменив порядок разыменования и преобразования к int* так: *(int*)v

92.

main()
{
  int i=i++, j=j++, k=k++;
  printf("%d%d%d",i,j,k);
}

  • Answer: Garbage values.
  • Explanation. An identifier is available to use in program code from the point of its declaration. So expressions such as i = i++ are valid statements. The i, j and k are automatic variables and so they contain some garbage value. Garbage in is garbage out (GIGO).

93.

main()
{
  static int i=i++, j=j++, k=k++;
  printf("i = %d j = %d k = %d", i, j, k);
}

  • Answer: i = 1 j = 1 k = 1
  • Explanation. Since static variables are initialized to zero by default.

94.

main()
{
  while(1) {
    if (printf("%d",printf("%d")))
      break;
    else
      continue;
  }
}

  • Answer: Garbage values
  • Explanation. The inner printf executes first to print some garbage value. The printf returns no of characters printed and this value also cannot be predicted. Still the outer printf prints something and so returns a non-zero value. So it encounters the break statement and comes out of the while statement.

95.

main()
{
  unsigned int i=10;
  while(i-- >= 0)
    printf("%u ",i);
}

  • Answer: 10 9 8 7 6 5 4 3 2 1 0 65535 65534…..
  • Explanation. Since i is an unsigned integer it can never become negative. So the expression i-- >=0 will always be true, leading to an infinite loop.
  • Комментарий. Два последних числа получены в предположении, что размер int 2 байта.

96.

main()
{
  int x,y=2,z,a;
  if (x=y%2) z=2;
  a=2;
  printf("%d %d ",z,x);
}

  • Answer: Garbage-value 0
  • Explanation. The value of y%2 is 0. This value is assigned to x. The condition reduces to if (x) or in other words if(0) and so z goes uninitialized.
  • Thumb Rule: Check all control paths to write bug free code.

97.

main()
{
  int a[10];
  printf("%d",*a+1-*a+3);
}

  • Answer: 4
  • Explanation: *a and -*a cancels out. The result is as simple as 1 + 3 = 4

98.

#define prod(a,b) a*b

main() 
{
  int x=3,y=4;
  printf("%d",prod(x+2,y-1));
}

  • Answer: 10
  • Explanation. The macro expands and evaluates to as: x+2*y-1 => x+(2*y)-1 => 10

99.

main()
{
  unsigned int i=65000;
  while(i++!=0);
  printf("%d",i);
}

  • Answer: 1
  • Explanation. Note the semicolon after the while statement. When the value of i becomes 0 it comes out of while loop. Due to post-increment on i the value of i while printing is 1.

100.

main()
{
  int i=0;
  while(+(+i--)!=0)
    i-=i++;
  printf("%d",i);
}

  • Answer: -1
  • Explanation. Unary + is the only dummy operator in C. So it has no effect on the expression and now the while loop is, while(i--!=0) which is false and so breaks out of while loop. The value –1 is printed due to the post-decrement operator.

101.

main()
{
  float f=5,g=10;
  enum{i=10,j=20,k=50};
  printf("%d\n",++k);
  printf("%f\n",f<<2);
  printf("%lf\n",f%g);
  printf("%lf\n",fmod(f,g)); 
}

  • Answer:
    • Line no 5: Error: Lvalue required
    • Line no 6: Cannot apply leftshift to float
    • Line no 7: Cannot apply mod to float
  • Explanation:
    • Enumeration constants cannot be modified, so you cannot apply ++.
    • Bit-wise operators and % operators cannot be applied on float values.
    • fmod() is to find the modulus values for floats as % operator is for ints.

102.

main()
{
  signed char i=0;
  for(;i>=0;i++);
  printf("%d\n",i);
}

  • Answer: -128
  • Explanation. Notice the semicolon at the end of the for loop. The initial value of the i is set to 0. The inner loop executes to increment the value from 0 to 127 (the positive range of char) and then it rotates to the negative value of -128. The condition in the for loop fails and so comes out of the for loop. It prints the current value of i that is -128.

103.

main()
{
  unsigned char i=0;
  for(;i>=0;i++);
  printf("%d\n",i);
}

  • Answer: infinite loop
  • Explanation. The difference between the previous question and this one is that the char is declared to be unsigned. So the i++ can never yield negative value and i>=0 never becomes false so that it can come out of the for loop.

104.

main()
{
  char i=0;
  for(;i>=0;i++);
  printf("%d\n",i);       
}

  • Answer: Behavior is implementation dependent.
  • Explanation. The detail if the char is signed/unsigned by default is implementation dependent. If the implementation treats the char to be signed by default the program will print –128 and terminate. On the other hand if it considers char to be unsigned by default, it goes to infinite loop.

105.

typedef enum errorType {warning, error, exception,} error;
main()
{
  error g1;
  g1=1; 
  printf("%d",g1);
}

  • Answer: Compiler error: Multiple declaration for error
  • Explanation. The name error is used in the two meanings. One means that it is a enumerator constant with value 1. The another use is that it is a type name (due to typedef) for enum errorType. Given a situation the compiler cannot distinguish the meaning of error to know in what sense the error is used:
    error g1;
    g1=error;
    // which error it refers in each case?
    When the compiler can distinguish between usages then it will not issue error (in pure technical terms, names can only be overloaded in different namespaces).
  • Note: the extra comma in the declaration, enum errorType{warning, error, exception,} is not an error. An extra comma is valid and is provided just for programmer’s convenience.
  • Комментарий. Кроме указанной ошибки, gcc находит ещё некорректное преобразование из «int» в тип «errror».

106.

typedef struct error {int warning, error, exception;} error;
main()
{
  error g1;
  g1.error =1; 
  printf("%d",g1.error);
}

  • Answer: 1
  • Explanation. The three usages of name errors can be distinguishable by the compiler at any instance, so valid (they are in different namespaces).

1. typedef struct **error** {int warning, error, exception;} error;

This error can be used only by preceding the error by struct keyword as in: struct error someError;

2. typedef struct error {int warning, **error**, exception;} error;

This can be used only after . (dot) or -> (arrow) operator preceded by the variable name as in:

 g1.error=1; 
 printf("%d",g1.error);

3. typedef struct error {int warning, error, exception;} **error**;

This can be used to define variables without using the preceding struct keyword as in: error g1; Since the compiler can perfectly distinguish between these three usages, it is perfectly legal and valid.

  • Note. This code is given here to just explain the concept behind. In real programming don’t use such overloading of names. It reduces the readability of the code. Possible doesn’t mean that we should use it!

107.

#ifdef something
int some=0;
#endif

main()
{
  int thing = 0;
  printf("%d %d\n", some ,thing);
}

  • Answer: Compiler error : undefined symbol some
  • Explanation. This is a very simple example for conditional compilation. The name something is not already known to the compiler making the declaration int some = 0; effectively removed from the source code.

108.

#if something == 0
int some=0;
#endif

main()
{
  int thing = 0;
  printf("%d %d\n", some ,thing);
}

  • Answer: 0 0
  • Explanation. This code is to show that preprocessor expressions are not the same as the ordinary expressions. If a name is not known the preprocessor treats it to be equal to zero.

109.

main()
{
  int arr2D[3][3];
  printf("%d\n", ((arr2D == *arr2D) && (*arr2D == arr2D[0])) );
}

  • Answer: 1
  • Explanation. This is due to the close relation between the arrays and pointers. N dimensional arrays are made up of (N-1) dimensional arrays.

The name arr2D refers to the beginning of all the 3 arrays. *arr2D refers to the start of the first 1D array (of 3 integers) that is the same address as arr2D. So the expression (arr2D == *arr2D) is true (1).

Similarly, *arr2D is nothing but *(arr2D + 0), adding a zero doesn’t change the value/meaning. Again arr2D[0] is the another way of telling *(arr2D + 0). So the expression (*(arr2D + 0) == arr2D[0]) is true (1).

Since both parts of the expression evaluates to true the result is true(1) and the same is printed.

  • Комментарий. Компилятор жалуется на первое из условий внутри printf() ((arr2D == *arr2D)): ошибка: comparison between distinct pointer types «int (*)[3]» and «int*» lacks a cast

Объяснение звучит логично, что можно подтвердить, запустив такой код:

#include <stdio.h>

int main()
{
  int arr2D[3][3] = {{0,1,2},{3,4,5},{6,7,8}};
  printf("%p\n", arr2D);

  int i,j;
  for (i = 0; i < 3; i++) {
    for (j = 0; j < 3; j++)
      printf("%p ", &arr2D[i][j]);
    printf("\n");
  }

  return 0;
}

Однако, из-за попытки приравнять указатели на различные типы, компилятор его не пропустит.

110.

main()
{
  if(~0 == (unsigned int)-1)
    printf("You can answer this if you know how values are represented in memory");
}

  • Answer: You can answer this if you know how values are represented in memory.
  • Explanation. ~ (tilde operator or bit-wise negation operator) operates on 0 to produce all ones to fill the space for an integer. –1 is represented in unsigned value as all 1’s and so both are equal.

111.

int swap(int *a,int *b)
{
 *a=*a+*b; *b=*a-*b; *a=*a-*b;
}

main()
{
  int x=10,y=20;
  swap(&x,&y);
  printf("x = %d y = %d\n",x,y);
}

  • Answer: x = 20 y = 10
  • Explanation. This is one way of swapping two values. Simple checking will help understand this.

112.

main()
{
  int i=5;
  printf("%d",++i++);
} 

  • Answer: Compiler error: Lvalue required in function main
  • Explanation: ++i yields an rvalue. For postfix ++ to operate an lvalue is required.

113.

int aaa() { printf("Hi"); }
int bbb() { printf("hello"); }
int ccc() { printf("bye"); }

main()
{
    int (*ptr[3])();
    ptr[0] = aaa;
    ptr[1] = bbb;
    ptr[2] = ccc;
    ptr[2]();
}

  • Answer: bye
  • Explanation. int (*ptr[3])() says that ptr is an array of pointers to functions that takes no arguments and returns the type int. By the assignment ptr[0] = aaa; it means that the first function pointer in the array is initialized with the address of the function aaa. Similarly, the other two array elements also get initialized with the addresses of the functions bbb and ccc. Since ptr[2] contains the address of the function ccc, the call to the function ptr2 is same as calling ccc(). So it results in printing "bye".

114.

int main()
{
  int i=5;
  printf("%d",i=++i==6);
}

  • Answer: 1
  • Explanation. The expression can be treated as i = (++i==6), because == is of higher precedence than = operator. In the inner expression, ++i is equal to 6 yielding true(1). Hence the result.

115.

main()
{
  char p[ ]="%d\n";
  p[1] = 'c';
  printf(p,65);
}

  • Answer: A
  • Explanation. Due to the assignment p[1] = ‘c’ the string becomes, "%c\n". Since this string becomes the format string for printf and ASCII value of 65 is ‘A’, the same gets printed.

116.

main()
{
  while (strcmp("some","some\0")) 
    printf("Strings are not equal\n");
}

  • Answer: No output
  • Explanation. Ending the string constant with \0 explicitly makes no difference. So "some" and "some\0" are equivalent. So, strcmp returns 0 (false) hence breaking out of the while loop.

117.

main()
{
  char str1[] = {'s','o','m','e'};
  char str2[] = {'s','o','m','e','\0'};
  while (strcmp(str1,str2)) 
    printf("Strings are not equal\n");
}

  • Answer:
    "Strings are not equal"
    "Strings are not equal"
    ….
  • Explanation. If a string constant is initialized explicitly with characters, ‘\0’ is not appended automatically to the string. Since str1 doesn’t have null termination, it treats whatever the values that are in the following positions as part of the string until it randomly reaches a ‘\0’. So str1 and str2 are not the same, hence the result.

118.

main()
{
  int i = 3;
  for (; i++=0; ) printf("%d",i);
}

  • Answer: Compiler Error: Lvalue required.
  • Explanation. The increment operators return rvalues and hence it cannot appear on the left hand side of an assignment operation.

119.

main()
{
  int *mptr, *cptr;
  mptr = (int*) malloc(sizeof(int));
  printf("%d",*mptr);
  int *cptr = (int*) calloc(sizeof(int),1);
  printf("%d",*cptr);
}

  • Answer: garbage-value 0
  • Explanation. The memory space allocated by malloc is uninitialized, whereas calloc returns the allocated memory space initialized to zeros.

120.

main()
{
  static int i;
  while (i<=10)
    (i>2)?i++:i--;
  printf("%d", i);
}

  • Answer: 32767
  • Explanation. Since i is static it is initialized to 0. Inside the while loop the conditional operator evaluates to false, executing i--. This continues till the integer value rotates to positive value (32767). The while condition becomes false and hence, comes out of the while loop, printing the i value.
  • Комментарий. Точнее, результат будет равен значению INT_MAX (limits.h) для вашей системы.

121.

main()
{
  int i=10,j=20;
  j = i, j ? (i,j) ? i : j : j;
  printf("%d %d",i,j);
}

  • Answer: 10 10
  • Explanation. The Ternary operator ( ? : ) is equivalent for if-then-else statement. So the question can be written as:
if(i,j){
  if(i,j)
    j = i;
  else
    j = j;          
}
else
  j = j;        

122.

main()
{
  int i=5,j=10;
  i = i &= j && 10;
  printf("%d %d",i,j);
}

  • Answer: 1 10
  • Explanation. The expression can be written as i=(i&=(j&&10)); The inner expression (j&&10) evaluates to 1 because j==10. i is 5. i = 5&1 is 1. Hence the result.

123.

main()
{
  int i=4, j=7;
  j = j || i++ && printf("YOU CAN");
  printf("%d %d", i, j);
}

  • Answer: 4 1
  • Explanation. The boolean expression needs to be evaluated only till the truth value of the expression is not known. j is not equal to zero itself means that the expression’s truth value is 1. Because it is followed by || and true || (anything) => true where (anything) will not be evaluated. So the remaining expression is not evaluated and so the value of i remains the same. Similarly when && operator is involved in an expression, when any of the operands become false, the whole expression’s truth value becomes false and hence the remaining expression will not be evaluated. false && (anything) => false where (anything) will not be evaluated.

124.

main()
{
  register int a=2;
  printf("Address of a = %d",&a);
  printf("Value of a   = %d",a);
}

  • Answer: Complier Error: '&' on register variable
  • Rule to Remember: & (address of) operator cannot be applied on register variables.

125.

main()
{
  float i=1.5;
  switch(i)
  {
  case 1: printf("1");
  case 2: printf("2");
  default : printf("0");
  }
}

  • Answer: Compiler Error: switch expression not integral
  • Explanation. Switch statements can be applied only to integral types.

126.

main()
{   
  extern i;
  printf("%d\n",i);
  {
    int i=20;
    printf("%d\n",i);
  }
}

  • Answer: Linker Error: Unresolved external symbol i
  • Explanation. The identifier i is available in the inner block and so using extern has no use in resolving it.

127.

main()
{
  int a=2,*f1,*f2;
  f1=f2=&a;
  *f1+=*f2+=a+=2.5;
  printf("\n%d %d %d",a,*f1,*f2);
}

  • Answer: 16 16 16
  • Explanation. f1 and f2 both refer to the same memory location a. So changes through f1 and f2 ultimately affects only the value of a.
  • Комментарий. В моём случае ответ 8 8 8 Похоже, компилятор делает все добавления к исходной величине: a = 2. Заявленный составителями ответ получится при разбиении операций '+=' на отдельные строки:
a+=2.5;
*f1+=a;
*f2+=*f1;

128.

main()
{
  char *p="GOOD";
  char a[]="GOOD";
  printf("\n sizeof(p) = %d, sizeof(*p) = %d, strlen(p) = %d", sizeof(p), sizeof(*p), strlen(p));
  printf("\n sizeof(a) = %d, strlen(a) = %d", sizeof(a), strlen(a));
}

  • Answer:
    sizeof(p) = 2, sizeof(*p) = 1, strlen(p) = 4
    sizeof(a) = 5, strlen(a) = 4
  • Explanation:
    sizeof(p) => sizeof(char*) => 2
    sizeof(*p) => sizeof(char) => 1
    Similarly,
    sizeof(a) => size of the character array => 5
    When sizeof operator is applied to an array it returns the sizeof the array and it is not the same as the sizeof the pointer variable. Here the sizeof(a) where a is the character array and the size of the array is 5 because the space necessary for the terminating NULL character should also be taken into account.

129.

#define DIM( array, type) sizeof(array)/sizeof(type)
main()
{
  int arr[10];
  printf("The dimension of the array is %d", DIM(arr, int));    
}

  • Answer: 10
  • Explanation. The size of integer array of 10 elements is 10 * sizeof(int). The macro expands to sizeof(arr)/sizeof(int) => 10 * sizeof(int) / sizeof(int) => 10.

130.

int DIM(int array[]) 
{
  return sizeof(array)/sizeof(int);
}

main()
{
  int arr[10];
  printf("The dimension of the array is %d", DIM(arr));
}

  • Answer: 1
  • Explanation. Arrays cannot be passed to functions as arguments and only the pointers can be passed. So the argument is equivalent to int * array (this is one of the very few places where [] and * usage are equivalent). The return statement becomes, sizeof(int *)/ sizeof(int) that happens to be equal in this case.

131.

main()
{
  static int a[3][3]={1,2,3,4,5,6,7,8,9};
  int i,j;
  static *p[]={a,a+1,a+2};
  for(i=0;i<3;i++)
  {
    for(j=0;j<3;j++)
      printf("%d\t%d\t%d\t%d\n",*(*(p+i)+j),*(*(j+p)+i),*(*(i+p)+j),*(*(p+j)+i));
  }
}

  • Answer:
    1 1 1 1
    2 4 2 4
    3 7 3 7
    4 2 4 2
    5 5 5 5
    6 8 6 8
    7 3 7 3
    8 6 8 6
    9 9 9 9
  • Explanation: *(*(p+i)+j) is equivalent to p[i][j].

132.

main()
{
  void swap();
  int x=10, y=8;
  swap(&x,&y);
  printf("x=%d y=%d",x,y);
}

void swap(int *a, int *b)
{
  *a ^= *b, *b ^= *a, *a ^= *b; 
}

  • Answer: x=8 y=10
  • Explanation. Using ^ like this is a way to swap two variables without using a temporary variable and that too in a single statement.

Inside main(), void swap(); means that swap is a function that may take any number of arguments (not no arguments) and returns nothing. So this doesn’t issue a compiler error by the call swap(&x,&y); that has two arguments.

This convention is historically due to pre-ANSI style (referred to as Kernighan and Ritchie style) style of function declaration. In that style, the swap function will be defined as follows,

void swap()
int *a, int *b
{
   *a ^= *b,  *b ^= *a, *a ^= *b;
}

where the arguments follow the (). So naturally the declaration for swap will look like, void swap() which means the swap can take any number of arguments.

  • Комментарий. Несмотря на олдскульный стиль, здесь есть любопытный пример swap'а.

133.

main()
{
  int i = 257;
  int *iPtr = &i;
  printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}

  • Answer: 1 1
  • Explanation. The integer value 257 is stored in the memory as, 00000001 00000001, so the individual bytes are taken by casting it to char * and get printed.

134.

main()
{
  int i = 258;
  int *iPtr = &i;
  printf("%d %d", *((char*)iPtr), *((char*)iPtr+1) );
}

  • Answer: 2 1
  • Explanation. The integer value 257 can be represented in binary as, 00000001 00000001. Remember that the Intel machines are ‘little-endian’ machines. Little-endian means that the lower order bytes are stored in the higher memory addresses and the higher order bytes are stored in lower addresses. So, the integer value 258 is stored in little-endian machine memory as 00000010 00000001.

135.

main()
{
  int i=300;
  char *ptr = &i; // (2)
  *++ptr=2;
  printf("%d",i);
}

  • Answer: 556
  • Explanation. The integer value 300 in binary notation is: 00000001 00101100. It is stored in memory (small-endian) as: 00101100 00000001. Result of the expression *++ptr = 2 makes the memory representation as: 00101100 00000010. So the integer corresponding to it is 00000010 00101100 => 556.
  • Комментарий. По дороге получим в строке (2) предупреждение компилятора об инициализации указателем несовместимого типа.

136.

main()
{
  char * str = "hello";
  char * ptr = str;
  char least = 127;
  while (*ptr++)
    least = (*ptr<least ) ? *ptr : least;
  printf("%d",least);
}

  • Answer: 0
  • Explanation. After ‘ptr’ reaches the end of the string the value pointed by ‘str’ is ‘\0’. So the value of ‘str’ is less than that of ‘least’. So the value of ‘least’ finally is 0.

137.

main()
{
  struct date;

  struct student
  {
    char name[30];
    struct date dob;
  } stud;

  struct date
  {
    int day,month,year;
  };

  scanf("%s%d%d%d", stud.name, &stud.dob.day, &stud.dob.month, &stud.dob.year);
}

  • Answer: Compiler Error: Undefined structure date
  • Explanation. Only declaration of struct date is available inside the structure definition of ‘student’ but to have a variable of type struct date the definition of the structure is required.
  • Комментарий. Сообщение об ошибке такое: поле «dob» имеет неполный тип

138.

main()
{
  int *i = 0x400; // i points to the address 400
  *i = 0;         // set the value of memory location pointed by i;
}

  • Answer: Undefined behavior
  • Explanation. The second statement results in undefined behavior because it points to some location whose value may not be available for modification. This type of pointer in which the non-availability of the implementation of the referenced location is known as 'incomplete type'.
  • Комментарий. Мудрый gcc в первой строке main() выдаёт: ошибка: некорректное преобразование из «int» в «int*» [-fpermissive]

139.

main()
{
  printf("sizeof (void *) = %d \n", sizeof(void *));
  printf("sizeof (int *)    = %d \n", sizeof(int *));
  printf("sizeof (double *)  = %d \n", sizeof(double *));
  printf("sizeof (struct unknown *) = %d \n", sizeof(struct unknown *));
}

  • Answer:
    sizeof (void *) = 2
    sizeof (int *) = 2
    sizeof (double *) = 2
    sizeof (struct unknown *) = 2
  • Explanation. The pointer to any type is of same size.

140.

main()
{
  int i=10, j=2;
  int *ip = &i, *jp = &j;
  int k = *ip/*jp;
  printf("%d",k);
}

  • Answer: Compiler Error: "Unexpected end of file in comment started in line 5".
  • Explanation. The programmer intended to divide two integers, but by the "maximum munch" rule, the compiler treats the operator sequence / and * as /* which happens to be the starting of comment. To force what is intended by the programmer,
int k = *ip/ *jp;   
// give space explicity separating / and * 
//or
int k = *ip/(*jp);
// put braces to force the intention will solve the problem.
  • Комментарий. Как видно, даже плагин отвечающий за отображение кода на сайте, сразу выделил указанное место как комментарий.

141.

main()
{
  char ch;
  for(ch=0; ch<=127; ch++)
    printf("%c   %d \n", ch, ch);
}

  • Answer: Implementation dependent
  • Explanation. The char type may be signed or unsigned by default. If it is signed then ch++ is executed after ch reaches 127 and rotates back to -128. Thus ch is always smaller than 127.
  • Комментарий. В моем случае тип знаковый, получился бесконечный цикл.

142.

main()
{
  char a[4]="HELLO";
  printf("%s",a);
}   

  • Answer: Compiler error: Too many initializers
  • Explanation. The array a is of size 4 but the string constant requires 6 bytes to get stored.
  • Комментарий. Компилятор ограничился предупреждением: initializer-string for array of chars is too long а при выполнении вывелось то, под что было отведено место HELL

143.

main()
{   
  char a[4]="HELL";
  printf("%s",a);
}

  • Answer: HELL%@!~@!@???@~~!
  • Explanation. The character array has the memory just enough to hold the string "HELL" and doesn't have enough space to store the terminating null character. So it prints the HELL correctly and continues to print garbage values till it accidentally comes across a NULL character.
  • Комментарий. В моём случае дело ограничилось печатью HELL.

144.

main()
{ 
  int a=10,*j;
  void *k;
  j=k=&a;
  j++;
  k++;
  printf("\n %u %u ",j,k);
}

  • Answer: Compiler error: Cannot increment a void pointer
  • Explanation. Void pointers are generic pointers and they can be used only when the type is not known and as an intermediate address storage type. No pointer arithmetic can be done on it and you cannot apply indirection operator (*) on void pointers.

145.

char *someFun()
{
  char *temp = "string constant";
  return temp;
}

main()
{
  puts(someFun());
}

  • Answer: string constant
  • Explanation. The program suffers no problem and gives the output correctly because the character constants are stored in code/data area and not allocated in stack, so this doesn’t lead to dangling pointers. *Комментарий. Да, есть такая любопытная особенность строковых литералов.

146.

char *someFun1()
{
  char temp[ ] = "string";
  return temp;
}

char *someFun2()
{
  char temp[ ] = {s,t,r,i,n,g};
  return temp;
}

main()
{
  puts(someFun1());
  puts(someFun2());
}

  • Answer: Garbage values.
  • Explanation. Both the functions suffer from the problem of dangling pointers. In someFun1() temp is a character array and so the space for it is allocated in heap and is initialized with character string "string". This is created dynamically as the function is called, so is also deleted dynamically on exiting the function so the string data is not available in the calling function main() leading to print some garbage values. The function someFun2() also suffers from the same problem but the problem can be easily identified in this case.

147.

main()
{
  char *p = "ayqm";
  printf("%c",++*(p++));
}

  • Answer: Run-rime error
  • Explanation. The p is a character pointer point to a constant string. We can't modified it and in above program we try to modified at ++*(p++), so its lead to be error. ++*p will cause error too.


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