Alex Rivera | Logout

Sign of a floating point number

Asked 2010-11-20T22:00:33.777
10

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.

Edit
Report

2 Answers

2

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.

answered 2010-11-20T22:16:36.660
0

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;
}
answered 2012-06-04T01:05:56.297

Your Answer