Checking whether a number is positive or negative using bitwise operators

Question

I can check whether a number is odd/even using bitwise operators. Can I check whether a number is positive/zero/negative without using any conditional statements/operators l

Answer 1

#include<stdio.h>

void main()
{
    int n;  // assuming int to be 32 bit long

    //shift it right 31 times so that MSB comes to LSB's position
    //and then and it with 0x1
    if ((n>>31) & 0x1 == 1) {
        printf("negative number\n");
    } else {
        printf("positive number\n");
    }

    getch();
}

Answer 2

// if (x < 0) return -1
// else if (x == 0) return 0
// else return 1
int sign(int x) {
  // x_is_not_zero = 0 if x is 0 else x_is_not_zero = 1
  int x_is_not_zero = (( x | (~x + 1)) >> 31) & 0x1;
  return (x & 0x01 << 31) >> 31 | x_is_not_zero; // for minux x, don't care the last operand 
}

Here's exactly what you waht!

Answer 3

Or, you could use signbit() and the work's done for you.

I'm assuming that under the hood, the math.h implementation is an efficient bitwise check (possibly solving your original goal).

Reference: http://en.cppreference.com/w/cpp/numeric/math/signbit

Answer 4

Here is an update related to C++11 for this old question. It is also worth considering std::signbit.

On Compiler Explorer using gcc 7.3 64bit with -O3 optimization, this code

bool s1(double d)
{
    return d < 0.0;
}

generates

s1(double):
  pxor xmm1, xmm1
  ucomisd xmm1, xmm0
  seta al
  ret

And this code

bool s2(double d)
{
    return std::signbit(d);
}

generates

s2(double):
  movmskpd eax, xmm0
  and eax, 1
  ret

You would need to profile to ensure that there is any speed difference, but the signbit version does use 1 less opcode.