Sign of a floating point number

Question

Is there an easy way to determine the sign of a floating point number?

I experimented and came up with this:

#include <iostream>

int main(int argc, char** argv)
{
 union
 {
  float f;
  char c[4];
 };

 f = -0.0f;
 std::cout << (c[3] & 0x10000000) << "\n";

 std::cin.ignore();
 std::cin.get();
 return 0;
}

where (c[3] & 0x10000000) gives a value > 0 for a negative number but I think this requires me to make the assumptions that:

• The machine's bytes are 8 bits big
• a float point number is 4 bytes big?
• the machine's most significant bit is the left-most bit (endianness?)

Please correct me if any of those assumptions are wrong or if I have missed any.

Answer 1

1) sizeof(int) has nothing to do with it.

2) assuming CHAR_BIT == 8, yes.

3) we need MSB for that, but endianness affects only byte order, not bit order, so the bit we need to check is c[0]&0x80 for big endianness, or c[3]&0x80 for little, so it would be better to declare union with an uint32_t and checking with 0x80000000.

This trick have sense only for non-special memory operands. Doing it to a float value that is in XMM or x87 register will be slower than direct approach. Also, it doesn't treat the special values like NaN or INF.

Answer 2

I've got this from http://www.cs.uaf.edu/2008/fall/cs441/lecture/10_07_float.html try this:

/* IEEE floating-point number's bits:  sign  exponent   mantissa */
struct float_bits {
    unsigned int fraction:23; /**< Value is binary 1.fraction ("mantissa") */
    unsigned int exp:8; /**< Value is 2^(exp-127) */
    unsigned int sign:1; /**< 0 for positive, 1 for negative */
};

/* A union is a struct where all the fields *overlap* each other */
union float_dissector {
    float f;
    struct float_bits b;
};

int main() {
    union float_dissector s;
    s.f = 16;
    printf("float %f  sign %u  exp %d  fraction %u",s.f, s.b.sign,((int)s.b.exp - 127),s.b.fraction);
    return 0;
}

Answer 3

Coming to this late, but I thought of another approach.

If you know your system uses IEEE754 floating-point format, but not how big the floating-point types are relative to the integer types, you could do something like this:

bool isFloatIEEE754Negative(float f)
{
    float d = f;
    if (sizeof(float)==sizeof(unsigned short int)) {
        return (*(unsigned short int *)(&d) >> (sizeof(unsigned short int)*CHAR_BIT - 1) == 1);
    }
    else if (sizeof(float)==sizeof(unsigned int)) {
        return (*(unsigned int *)(&d) >> (sizeof(unsigned int)*CHAR_BIT - 1) == 1);
    }
    else if (sizeof(float)==sizeof(unsigned long)) {
        return (*(unsigned long *)(&d) >> (sizeof(unsigned long)*CHAR_BIT - 1) == 1);
    }
    else if (sizeof(float)==sizeof(unsigned char)) {
        return (*(unsigned char *)(&d) >> (sizeof(unsigned char)*CHAR_BIT - 1) == 1);
    }
    else if (sizeof(float)==sizeof(unsigned long long)) {
        return (*(unsigned long long *)(&d) >> (sizeof(unsigned long long)*CHAR_BIT - 1) == 1);
    }
    return false; // Should never get here if you've covered all the potential types!
}

Essentially, you treat the bytes in your float as an unsigned integer type, then right-shift all but one of the bits (the sign bit) out of existence. '>>' works regardless of endianness so this bypasses that issue.

If it's possible to determine pre-execution which unsigned integer type is the same length as the floating point type, you could abbreviate this:

#define FLOAT_EQUIV_AS_UINT unsigned int // or whatever it is

bool isFloatIEEE754Negative(float f)
{
    float d = f;
    return (*(FLOAT_EQUIV_AS_UINT *)(&d) >> (sizeof(FLOAT_EQUIV_AS_UINT)*CHAR_BIT - 1) == 1);
}

This worked on my test systems; anyone see any caveats or overlooked 'gotchas'?

Answer 4

Try

float s = copysign(1, f);

from <math.h>

Another helpful thing may be #including <ieee754.h>, if it's available on your system/compiler.

Answer 5

Why not if (f < 0.0)?

Answer 6

Use signbit() from math.h.