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Give a method to count the number of ones in a 32 bit number?
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Answer / ataraxic
int i = 0;
while (x) {
x &= x-1;
++i;
};
printf("number of ones is %d\n", i);
| Is This Answer Correct ? | 1 Yes | 0 No |
#include<stdio.h>
#include<conio.h>
void main()
{
unsigned i;
int j=0,count=0;;
printf("Enter the number :");
scanf("%ld",&i);
while(j<=31)
{
if(!(((i>>j)&1)^1))
count++;
j++;
}
printf("\nnumber of 1's in ur number is : %d",count);
getch();
}
thank u
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / jayaprakash
#include<stdio.h>
#include<conio.h>
main()
{
int i;
int n;
int count=0;
int j;
int res=0;
clrscr();
printf("Enter the number:");
scanf("%d",&n);
for(j=15;j>=0;j--)
{ i=1;
i=i<<j;
res=i&n;
if(res!=0)
count++;
}
printf("\nNumber of ones is:%d",count);
getch();
}
| Is This Answer Correct ? | 0 Yes | 0 No |
Answer / jayaprakash
#include<stdio.h>
#include<conio.h>
main()
{
int i;
int n;
int count=0;
int j;
int res=0;
clrscr();
printf("Enter the number:");
scanf("%d",&n);
for(j=15;j>=0;j--)
{ i=1;
i=i<<j;
res=i&n;
if(res!=0)
count++;
}
printf("\nNumber of ones is:%d",count);
getch();
}
| Is This Answer Correct ? | 0 Yes | 0 No |
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