How to print 1 when any bit of a number equals 0?

Question

I need to print out the value 1 when any bit of a number equals 0, and am currently struggling with how I would go about doing this. I am not supposed to use any equality or inequality tests, and am basically restricted to just bit-level and logic operations. Any ideas?

int b = 15;
printf("prints 1 when any bit of a number equals 0: %d\n", //PRINT GOES HERE);

b = 10;
printf("prints 1 when any bit of a number equals 0: %d\n", //PRINT GOES HERE);

Answer 1

I do not understand your question but for two complement signed integer all bits are set for number -1

So you need to check if number is not equal -1

void foo(int a)
{
       printf("prints 1 when any bit of a number equals 0: %d\n", !((unsigned)a ^ (unsigned)-1));
}

int  main(void)
{
    foo(15);
}

But IMO it is too trivial. You probably want check starting from the first (most significant)set bit (so the 15 prints 0 but 14 prints 1 ):

unsigned mask(int a)
{
    int bit;
    if(!((unsigned)a ^ (unsigned)-1)) return -1;
    for(bit = sizeof(a)*CHAR_BIT - 1; bit; bit--)
    {
        if(((unsigned)a & (1U << bit))) break;
    }
    return (1U << (bit + 1)) - 1;
}

void foo(int a)
{
       printf("prints 1 when any bit of a number equals 0: %d\n", !!((a & mask(a)) ^ mask(a)));
}


int  main(void)
{
    foo(15);
}

Answer 2

You would want to create a mask with all 1's (~0x0) then XOR it with the number. If there are any 1's afterwords then at least one of the original bits was 0, and thus you can use &&1 to return 1 if there is at least one 1 and 0 if not.