Tricky Questions or Puzzles in C

1) WTAHT will be the output of the following Printf function.
  printf("%d",printf("%d",printf("%d",printf("%s","ILOVECPROGRAM"))));

Ans-ILOVECPROGRAM1321

The above printf line gives output like this because printf returns the number of character successfully written in the output.So the inner printf("%s","ILOVECPROGRAM") writes 13 characters to the output so the outer printf function will print 13 and as 13 is of 2 characters so the next outer printf function will print 2 and then next outer printf will print 1 as 2 is one character.So is the output 1321

2-Write code snippets to swap two variables in five different ways.

Answer:

a.      /* swapping using three variables*/ (Takes extra memory space)

Int a=5, b=10, c;

c=a;

a=b;

b=c;

b.      /* using arithmetic operators */

a=a+b;

b=a-b;

a=a-b;

c.       /* using bit-wise operators */

a=a^b;

b=b^a;

a=a^b;

Line	Operation	Value of a	Value of b
1	-	5	10	Initial values
2	a=a^b	15	10
3	b=a^a	15	5
4	a=a^b	10	5	values after swapping

d.      /* one line statement using bit-wise operators */ (most efficient)

a^=b^=a^=b;

The order of evaluation is from right to left. This is same as in approach (c) but the three statements are compounded into one statement.

e.       /* one line statement using arithmetic & assignment operators */

a=(a+b) - (b=a);

In the above axample, parenthesis operator enjoys the highest priority & the order of evaluation is from left to right. Hence (a+b) is evaluated first and replaced with 15. Then (b=a) is evaluated and the value of a is assigned to b, which is 5. Finally a is replaced with 15-5, i.e. 10. Now the two numbers are swapped.

3-  Find the maximum & minimum of two numbers in a single line without using any condition & loop.

Answer:

void main ()

{

int a=15, b=10;

printf (“ max = %d, min = %d”, ((a+b) + abs(a-b)) /2, ((a+b) – abs (a-b)) /2);

}

4.  How to print number from 1 to 100 without using conditional operators.

Answer:

void main ()

{

int  i=0;

while (100 – i++)

printf (“ %d”, i);

}

1.     #include <stdio.h>

2.     printf("%.0f", 2.89);

a. 2.890000
b. 2.89
c. 2
d.3 

 Answer

Answer:d

5-#include<stdio.h>

int main(){

    int i,j,k;

    for(i=0,j=2,k=1;i<=4;i++){

         printf("%d ",i+j+k);

    }

return 0;        

}

Output: 3 4 5 6 7