Advanced Test in C: The 0x10 Best Questions for C Programmers

• In all program, assume that the required header file/files has /have been included

1. Consider the following program:

#include<setjmp.h>
static jmp_buf  buf;

main()
{
  volatile  int b;
  b =3;

  if(setjmp(buf)!=0)  
  {
    printf("%d ", b);  
    exit(0);
  }
  b=5;
  longjmp(buf , 1);
}

2. Consider the following program:

main()
{
   struct node 
   {
     int a;
     int b;
     int c;     
   };
   struct node  s= { 3, 5,6 };
   struct node *pt = &s;
   printf("%d" ,  *(int*)pt);
}

3. Consider the following code segment:

int  foo ( int x , int  n)
{
  int val;
  val =1;
  
  if (n>0) 
  {
    if (n%2 == 1)  val = val *x;
    
    val = val * foo(x*x , n/2);
  }
  return val;
}

4. Consider the following program:

main()
{
  int  a[5] = {1,2,3,4,5};
  int *ptr =  (int*)(&a+1);

  printf("%d %d" , *(a+1), *(ptr-1) );

}

5. Consider the following program:

void foo(int [][3] );     

main()
{
  int a [3][3]= { { 1,2,3} , { 4,5,6},{7,8,9}};
  foo(a);
  printf("%d" , a[2][1]);
}

void foo( int b[][3])
{
  ++ b;
  b[1][1] =9;
}

6. Consider the following program:

main()
{
  int a, b,c, d;
  a=3;
  b=5;
  c=a,b;
  d=(a,b);

  printf("c=%d" ,c);
  printf("d=%d" ,d);

}

7. Consider the following program:

main()
{
  int a[][3] = { 1,2,3 ,4,5,6};
  int (*ptr)[3] =a;

  printf("%d %d "  ,(*ptr)[1], (*ptr)[2] );

  ++ptr;
  printf("%d %d"  ,(*ptr)[1], (*ptr)[2] );
}

8. Consider following function

int *f1(void)
{
  int x =10;
  return(&x);
}

int *f2(void)
{
  int*ptr;
  *ptr =10;
  return ptr;
}

int *f3(void)
{
  int *ptr;
  ptr=(int*) malloc(sizeof(int));
  return ptr;
}

9. Consider the following program:

main()
{
  int i=3;
  int j;

  j = sizeof(++i+ ++i);

  printf("i=%d j=%d", i ,j);
}

10. Consider the following program:

void f1(int *, int); 
void f2(int *, int); 
void(*p[2]) ( int *, int);

main()
{
  int a;
  int b;

  p[0] = f1;
  p[1] = f2;
  a=3;
  b=5;

  p[0](&a , b);
  printf("%d\t %d\t" , a ,b);

  p[1](&a , b);
  printf("%d\t %d\t" , a ,b);
}

void f1( int* p , int q)
{
  int tmp;
  tmp =*p;
  *p = q;
  q= tmp;
}

void f2( int* p , int q)
{
  int tmp;
  tmp =*p;
  *p = q;
  q= tmp;
}  

11. Consider the following program:

void e(int );   

main()
{
  int a;
  a=3;
  e(a);
}

void e(int n)
{
  if(n>0)
  {
    e(--n);
    printf("%d" , n);
    e(--n);
  }
}

12. Consider following declaration

typedef int (*test) ( float * , float*)
test tmp;

13. Consider the following program:

main()
{
  char *p;
  char buf[10] ={ 1,2,3,4,5,6,9,8};
  p = (buf+1)[5];
  printf("%d" , p);
}

14. Consider the following program:

Void f(char**);

main()
{
  char * argv[] = { "ab" ,"cd" , "ef" ,"gh", "ij" ,"kl" };
  f( argv );
}

void f( char **p )
{
  char* t;

  t= (p+= sizeof(int))[-1];

  printf( "%s" , t);
}

15. Consider the following program:

#include<stdarg.h>
int ripple ( int , ...);

main()
{
  int num;
  num = ripple ( 3, 5,7);
  printf( " %d" , num);
}

int ripple (int n, ...)
{
  int i , j;
  int k;  
  va_list p;

  k= 0;
  j = 1;
  va_start( p , n);     

  for (; j<n;  ++j) 
  {
    i =  va_arg( p , int);
    for (; i;    i &=i-1  )
      ++k;
  }
  return k;
}

16. Consider the following program:

int counter (int i)
{
  static int count =0;
  count = count +i;
  return (count );
}
main()
{
  int i , j;

  for (i=0; i <=5; i++)
    j = counter(i);
}

Answer With Detailed Explanation

Answer 1.

Answer 2.

Answer 3.

int  what ( int x , int  n)
{
  int val;
  int product;
  product =1;
  val =x;

  while(n>0)
  {
    if (n%2 == 1)  
      product = product*val;
    n = n/2; 
    val = val* val;
  }
}

(while n>0)  
{
  if  next most significant binary digit of  n( power)  is one
  then multiply accumulated product by current val  , 
  reduce n(power)  sequence by a factor of two using integer division .
  get next val by multiply current value of itself                   
}

Answer 4.

Answer 5.

Answer 6.

c=a,b;  / *yields c=a* /
d=(a,b); /* d =b  */

Answer 7.

Answer 8.

Answer 9.

Answer 10.

Answer 11.

Answer 12.

typedef char string

Answer 13.

Answer 14.

Answer 15.

(; i; i&=i-1) k++ /* count number of  1 bit in i *

let  i= 9 = 1001
     i-1  = 1000        
    (i-1) +1 = i
               1000
                 +1
              1 001

Answer 16.

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User Comments

Anonymous (Report as abusive)	Finland Excellect ..............
Abir Qaasem From DC (Report as abusive)	Interesting quiz -- #3 may have a problem The recursive code in #3 does not have return statement
Anonymous From bangalore (Report as abusive)	great c quiz what a quiz in c ?. wonderful
Infidel From Australia (Report as abusive)	Errors These are probably typos... The possible for answers for question 3 appear on the webpage as: (a) xn (b) x*n (c) nx (d) None of the above Since the answer is given as (a), it should read "(a) pow(x,n)" or with a superscript. I don't know what choice (c) is supposed to say. For question 13: main() { char *p; char buf[10] ={ 1,2,3,4,5,6,9,8}; p = (buf+1)[5]; printf("%d" , p); } My guess is that "p" is supposed to be declared as "char p" instead of as a pointer, but I may be wrong.
paul (Report as abusive)	great quiz on c most of the question are tricky and informative . good stuff for c programmers
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