Question

Here is a small Program to print powers of 2 till 8. But it is not quitting after 8. Please explain the reason.

#include <stdio.h>
#include <unistd.h>
int main(void)
{
        unsigned int i=1;
        while(1) {
                i = i<<1;
                printf("i = %d\n",i);
                if(i==(2^8))
                        break;
                sleep(1);
        }
        printf("Exited While loop.. \n");
        return 0;
}

The Loop is not Exiting when i = 2^8. My output is something like this:

i = 2
i = 4
i = 8
i = 16
i = 32
i = 64
i = 128
i = 256
i = 512 (Should have Exited here. But the program is continuing. Why?)
i = 1024
i = 2048
i = 4096....

EDIT :

Thanks for answering that ^ is an XOR operator. But now the below code is behaving strange. Please Explain.

#include <stdio.h>
int main(void)
{
        if((2)^8 == 1<<8) {
                printf("True.. \n");
        } else {
                printf("False..!!");
        }
        return 0;
}

The above function program prints true.

Answer 1

In C, the ^ operator means XOR (bitwise exclusive or).

To get 2 to the power of 8, you need to either use a loop (res *=2 in a loop), or round the pow function in math.h (note that the math.h function returns float – and therefore won’t be equal to the integer).

The simplest method is the bitwise shift, of course.

About the edit section:

Welcome to the wonderful world of operator precedence.
What happens is that == has higher precedence than ^, and therefore the conditional evaluates to 2^0, which is 2, which is true.

To make it work, you need to add parentheses:

if ( (2^8) == (1<<8) ) ...