Important C Questions and Answers

[Kalyan]

Important C Questions and Answers

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:
     The int ret(int ret), ie., the function name and the argument name can be the same.
     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.
void 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.
void main()
{
     unsigned giveit=-1;
     int gotit;
     printf("%u ",++giveit);
     printf("%u \n",gotit=--giveit);
}

Answer:
     0 65535

void 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.
void 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,
     1. Passing generic pointers to functions and returning such pointers.
     2. As a intermediate pointer type.
     3. Used when the exact pointer type will be known at a later point of time.
void 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).
void 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.
void 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.
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.
#include
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..n
Thumb Rule: Check all control paths to write bug free code