Binary representation of a number in C

Question

I came across this code for the binary representation of a number. I want to know the need for using !! in the code.

int main() {
    int n,i;
    unsigned flag = 1<<(sizeof(int) * 8 - 1);     

    printf("Input the number\n");
    scanf("%d",&n);     
    for(i=0;i<sizeof(int)*8;i++) {    
            printf("%d",!!(n & flag) );    
            n = n << 1;
    }
    return 0;
}

Answer 1

The flag used has only the MSB set and all other bits cleared, so that when you bitwise and it with number you can test the MSB in the number.

There are two outcomes of the bitwise anding:

• Zero - means the number had 0 in its MSB.
• Non-Zero - means the number had 1 in its MSB.

Now we need a way to map

Non-zero -> 1
Zero -> 0

so we use the double negation.

The same thing could have been done using:

for(i=0;i<sizeof(int)*8;i++) {

    (n & flag) ? printf("1"):printf("0");
    n = n << 1;
}

Answer 2

!! will convert any non-zero value to 1, and leave zero value as zero.

x = 0;
y = 50;
!x; // 1
!y; // 0
!!x; // 0
!!y; // 1

It is a poor man's bool cast.

Answer 3

I would write !!x less confusingly as x != 0.