Answer / lakshmi

void main()
{
int a;
scanf("%d",&a);

if((a&a-1)==0)
printf("Is a power of 2");
else
printf("Not a power of 2");
}

Answer / fallen angel

if(x&(x-1)==0)
then TRUE;
else
FALSE;

use the bitwise and operator

Answer / mayur shankariya

#include<stdio.h>
main()
{
int num;
printf("Enter Number
");
scanf("%d",&num);
(num & num - 1)? printf("Not
"):printf("Power of two
");
}

Answer / rarach

if( (a == 2 ) || ((a/2) %2 == 0 ) && a !=1)
TRUE
else
false

Answer / rohit

if((log(n)/log(2))/floor(log(n)/log(2))==1)
then TRUE;
else FALSE;

Answer / ramesh b penchal

#include<stdio.h>
main()
{
int n,i,d,m;
printf("Enter a number");
scanf("%d",&n);
m=n;
while(n>0)
{
d=n%10;
if(d!=0)
{
prntf("%d is not power of 2",m);
getch();
exit();
}
n=n/10;
}
prntf("%d is power of 2",m);
getch();
}

Answer / santhoo035

#include<iostream.h>
int main()
{
int n,i;
cout<<"Enter a number";
cin>>n;
for(i=1;i<n/2;i++)
{
if((2<<i)==n)
{
cout<<"The given no is power of 2";
break;
}
}
}

Answer / akshay rastogi

#include<conio.h>
void main()
{
int n,i=0,num=0;
printf("\n enter any number");
scanf("%d",&n);
while(num<=n)
{
i++;
num=2^i;
}
if(num==n)
printf("yes no. is power of 2");
else
printf("no.");
}

Answer / abc def

/*following expr evaluates to true if num is a power of
2.Else it's false. '&' - bitwise and.*/
(num == 1) || !(num & 1)

Answer / trailokya ranjan jena

#include<stdio.h>
void main()
{
int i,j;
clrscr();
printf("\n Enter a num");
scanf("%d",&i);
j=i;
for(;i%2==0;i/=2);
if(i==1)
printf("\n%d is power of 2",j);
else
printf("\n%d is not a power of 2");
getch();
}

main() { int c[ ]={2.8,3.4,4,6.7,5}; int j,*p=c,*q=c; for(j=0;j<5;j++) { printf(" %d ",*c); ++q; } for(j=0;j<5;j++){ printf(" %d ",*p); ++p; } }

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write the function. if all the character in string B appear in string A, return true, otherwise return false.

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Give a oneline C expression to test whether a number is a power of 2?

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