C Puzzles : Questions and Answers - Level 1

Questions

L1.Q1 : Write the output of the following program

#include 

#define	ABC	20
#define XYZ	10
#define XXX	ABC - XYZ

void main()
{
	int 	a;

	a = XXX * 10;

	printf("%d\n", a);
}

Solution for L1.Q1

L1.Q2 : Write the output of this program

#include 

#define calc(a, b)	(a * b) / (a - b)

void main()
{
	int a = 20, b = 10;

	printf("%d\n", calc(a + 4, b -2));
}

Solution for L1.Q2

L1.Q3 : What will be output of the following program ?

#include 

void main()
{
	int cnt = 5, a;

	do {
		a /= cnt;
	} while (cnt --);

	printf ("%d\n", a);
}

Solution for L1.Q3

L1.Q4 : Print the output of this program

#include 

void main()
{
	int  a, b, c, abc = 0;

	a = b = c = 40;

	if (c) {
		int abc;

		abc = a*b+c;
	}

	printf ("c = %d, abc = %d\n", c, abc);
}

Solution for L1.Q4

L1.Q5 : Print the output of this program

#include 

main()
{
	int k = 5;

	if (++k < 5 && k++/5 || ++k <= 8);

	printf("%d\n", k);
}

Solution for L1.Q5

L1.Q6 : What is the output of this program ?

#include 

void fn(int, int);

main()
{
	int 	a = 5;

	printf("Main :  %d %d\n", a++, ++a);

	fn(a, a++);
}

void fn(int a, int b)
{
	printf("Fn : a = %d \t b = %d\n", a, b);
}

Solution for L1.Q6

Answers

L1.A1
Solution for L1.Q1

a = xxx * 10  
which is => a = ABC - XYZ * 10
	 => a = 20 - 10 * 10
	 => a = 20 - 100
	 => a = -80

L1.A2
Solution for L1.Q2

Actual substitution is like this :

calc(20+4, 10 -2) is calculated as follows 

 	(20+4 * 10-2) / (20+4 - 10-2)
	(20+40-2) / 12
	58 / 12  = 4.8 
since  it is printed in %d the ans is 4

L1.A3
Solution for L1.Q3

This  problem  will compile  properly,  but it will give run
time error.  It will give divide-by-zero  error.  Look in to
the do loop portion

	do { a /= cnt; } while (cnt --);

when the 'cnt' value is 1, it is decremented in 'while
( cnt --)' and on next reference of 'cnt' it becomes zero.

	a /= cnt;  /* ie. a /= 0 */
which leads to divide-by-zero error.

L1.A4
Solution for L1.Q4

the result will be c = 40 and abc = 0;
because the scope of the variable  'abc' inside if(c) {..  }
is not valid out side that if (.)  { ..  }.

L1.A5
Solution for L1.Q5

The answer is 7.  The first condition ++k < 5 is checked and
it is false  (Now k = 6).  So, it checks  the 3rd  condition
(or condition ++k <= 8) and (now k = 7) it is true.  At this
point k value is  incremented by twice, hence the value of k
becomes 7.

L1.A6
Solution for L1.Q6

The  solution  depends  on  the   implementation  of  stack.
(Depends on OS) In some  machines the  arguments  are passed
from left to right to the  stack.  In this  case the  result
will be

	Main :  5 7 Fn :  7 7 

Other  machines  the  arguments  may be passed from right to
left to the stack.  In that case the result will be

	Main : 6 6
	Fn   : 8 7