Check if a number x is positive (x>0) by ONLY using bitwise operators in C

Question

isPositive - return true if x > 0, otherwise false

Example: isPositive(-1)

Legal ops:

Answer 1

You have another option here:

int is_positive = (0&x^((0^x)&-(0<x)));

It is just a (0 & min(0,x)).

Test here

Answer 2

if your working with a number system that uses the MSB as the signage bit, you can do:

int IsPositive(int x)
{
    return (((x >> 31) & 1) ^ 1) ^ !x;
}

Answer 3

Let's play with the sign bit: sign(~n) : 1 if n >= 0

To get rid of the case when n is 0: sign(~n + 1) : 1 if n > 0 or n = MIN_INT

So, we want the case when both functions return 1:

return ((~n & (~n + 1)) >> 31) & 1;

Answer 4

int isPositive(int x)
{
 return (!(x & 0x80000000) & !!x); 
}

Answer 5

Why not use XOR (^)?

Try this,

{
    return ((x>>31)&1)^(!!x); 
}

It can deal with the 0 case well.

Answer 6

int isPositive(int x) {
   return !((x&(1<<31)) | !x);
}

x&(1<<31 is to check if the number is negative.

!x is to check if the number is zero.

A number is positive if it's not negative and not zero.