Answer / banavathvishnu

int main()
{
int num1,num2;
int cnt = 0;
int temp1,temp2;
printf("enter 2 numbers \n");
scanf("%d %d",&num1,&num2);
while((num1!=0)||(num2!=0))
{
temp1= num1 & 0x01;
temp2 = num2 & 0x01;
if((temp1 ^ temp2)==1)
cnt++;
num1 = num1>>1;
num2 = num2>>1;
}
printf("difference is %d",cnt);
getch();
}

Answer / vadivelt

Hi Small Bug is there in my previous post.
That is corrected in the code written below.

#include<stdio.h>
#include<conio.h>

main()
{
int count = 0, n, i, Res = 0;
int a, b;
printf("GIVE 2 INPUT NO IS\n");
scanf("%d %d", &a, &b);
n = sizeof(int) * 8;
for(i = 0; i<n; i++)
{
Res = ((a >> i & 0x01) ^ (b >> i & 0x01)) ? 1 : 0;
if(Res == 1)
count++;
}
printf("BIT(S) DIFFERENCE: %d", count);
getch();
}

Answer / guest

The Question has to be corrected !!!. According to the
input given in the question, the bits difference should be
3.

#include<stdio.h>
#include<conio.h>

main()
{
int count = 0, n, i, Res = 0;
int a, b;
printf("GIVE 2 INPUT NOS\n");
scanf("%d%d", &a, &b);
n = sizeof(int);
for(i = 0; i<n; i++)
{
Res = ((a >> i & 0x01) & (b >> i & 0x01)) ? 1 : 0;
if(Res == 0)
count++;
}
printf("BIT(S) DIFFERENCE: %d", count);
getch();
}

Answer / subhash

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

printf("Enter 2 numbers:");
scanf("%d %d", &a, &b);

c = a ^ b; /* "c" holds bits set for different bits
in "a" and "b" *
/
while(c)
{
c &= (c-1);
count++;
}

printf("The different bits set:%d", count);
}

Just came across this question, felt worth sharing, so here it is I want you to make a C/C++ program that for any positive integer n will print all the positive integers from 1 up to it and then back again! Let's say n=5 I want the program to print: 1 2 3 4 5 4 3 2 1. Too easy you say? Okay then... You can ONLY USE: 1 for loop 1 printf/cout statement 2 integers( i and n) and as many operations you want. NO if statements, NO ternary operators, NO tables, NO pointers, NO functions!

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