## Comparing floating point numbers for equality.

Everyone who’s taken a architecture course (or messed around with scientific computing) knows that floating point numbers are not associative. That means mathematically:

$a*(b+c) \neq a*b+a*c$

Or, in layman’s terms:

The order of operations influences the result of the calculation

This implies that floating point calculations that mathematically give the same answer, does not necessarily produce exactly the same floating point number. So, when comparing two floating point results, using $a == b$ will not give the correct result.

You can attempt to remedy this by using the mathematical approach of allowing an absolute error metric:

$(a*b)^2 < E$

which does not account for the fact that floating point numbers are unequally distributed over the real number line. We can attempt to use a relative error metric:

$\frac{|a-b|}{b} < E$

but this does not take into account the difference between very small positive and negative numbers (including positive and negative zero, since floats have both).

So, from the very enlightening “comparing floats” article by Bruce Dawson, we try something quite different.

Floats can be lexographically ordered if you consider their bitstream to be signed-magnitude integers. We can exploit this fact to calculate exactly how many representable floating point numbers there are between two floats. So, for example, we can find that there is only one floating point number between 9,999.99999 and 10,000.00001 and use an error metric that states “I will consider floats to be equal if they are within $E$ representable floats of each other.

The details of this routine is in the comparing floats article, but I will mirror the code here:

// Usable AlmostEqual function

bool AlmostEqual2sComplement(float A, float B, int maxUlps)
{
// Make sure maxUlps is non-negative and small enough that the
// default NAN won't compare as equal to anything.
assert(maxUlps > 0 && maxUlps < 4 * 1024 * 1024);
int aInt = *(int*)&A;
// Make aInt lexicographically ordered as a twos-complement int
if (aInt < 0)
aInt = 0x80000000 - aInt;
// Make bInt lexicographically ordered as a twos-complement int
int bInt = *(int*)&B;
if (bInt < 0)
bInt = 0x80000000 - bInt;
int intDiff = abs(aInt - bInt);
if (intDiff <= maxUlps)
return true;
return false;

}

This has saved me huge amounts of headaches comparing CUDA and CPU generated results for our CUDA programming class, CS193G.