c++ - options gcc Different floating point result with optimization enabled - compiler bug?

3 Answers

Output should be: 4.5 4.6 That's what the output would be if you had infinite precision, or if you were working with a device that used a decimal-based rather than binary-based floating point representation. But, you aren't. Most computers use the binary IEEE floating point standard.

As Maxim Yegorushkin already noted in his answer, part of the problem is that internally your computer is using an 80 bit floating point representation. This is just part of the problem, though. The basis of the problem is that any number of the form n.nn5 does not have an exact binary floating representation. Those corner cases are always inexact numbers.

If you really want your rounding to be able to reliably round these corner cases, you need a rounding algorithm that addresses the fact that n.n5, n.nn5, or n.nnn5, etc. (but not n.5) is always inexact. Find the corner case that determines whether some input value rounds up or down and return the rounded-up or rounded-down value based on a comparison to this corner case. And you do need to take care that a optimizing compiler will not put that found corner case in an extended precision register.

See How does Excel successfully Rounds Floating numbers even though they are imprecise? for such an algorithm.

Or you can just live with the fact that the corner cases will sometimes round erroneously.

gcc floating point precision

The below code works on Visual Studio 2008 with and without optimization. But it only works on g++ without optimization (O0).

#include <cstdlib>
#include <iostream>
#include <cmath>

double round(double v, double digit)
    double pow = std::pow(10.0, digit);
    double t = v * pow;
    //std::cout << "t:" << t << std::endl;
    double r = std::floor(t + 0.5);
    //std::cout << "r:" << r << std::endl;
    return r / pow;

int main(int argc, char *argv[])
    std::cout << round(4.45, 1) << std::endl;
    std::cout << round(4.55, 1) << std::endl;

The output should be:


But g++ with optimization (O1 - O3) will output:


If I add the volatile keyword before t, it works, so might there be some kind of optimization bug?

Test on g++ 4.1.2, and 4.4.4.

Here is the result on ideone: http://ideone.com/Rz937

And the option I test on g++ is simple:

g++ -O2 round.cpp

The more interesting result, even I turn on /fp:fast option on Visual Studio 2008, the result still is correct.

Further question:

I was wondering, should I always turn on the -ffloat-store option?

Because the g++ version I tested is shipped with CentOS/Red Hat Linux 5 and CentOS/Redhat 6.

I compiled many of my programs under these platforms, and I am worried it will cause unexpected bugs inside my programs. It seems a little difficult to investigate all my C++ code and used libraries whether they have such problems. Any suggestion?

Is anyone interested in why even /fp:fast turned on, Visual Studio 2008 still works? It seems like Visual Studio 2008 is more reliable at this problem than g++?

To those who can't reproduce the bug: do not uncomment the commented out debug stmts, they affect the result.

This implies that the problem is related to the debug statements. And it looks like there's a rounding error caused by loading the values into registers during the output statements, which is why others found that you can fix this with -ffloat-store

Further question:

I was wondering, should I always turn on -ffloat-store option?

To be flippant, there must be a reason that some programmers don't turn on -ffloat-store, otherwise the option wouldn't exist (likewise, there must be a reason that some programmers do turn on -ffloat-store). I wouldn't recommend always turning it on or always turning it off. Turning it on prevents some optimizations, but turning it off allows for the kind of behavior you're getting.

But, generally, there is some mismatch between binary floating point numbers (like the computer uses) and decimal floating point numbers (that people are familiar with), and that mismatch can cause similar behavior to what your getting (to be clear, the behavior you're getting is not caused by this mismatch, but similar behavior can be). The thing is, since you already have some vagueness when dealing with floating point, I can't say that -ffloat-store makes it any better or any worse.

Instead, you may want to look into other solutions to the problem you're trying to solve (unfortunately, Koenig doesn't point to the actual paper, and I can't really find an obvious "canonical" place for it, so I'll have to send you to Google).

If you're not rounding for output purposes, I would probably look at std::modf() (in cmath) and std::numeric_limits<double>::epsilon() (in limits). Thinking over the original round() function, I believe it would be cleaner to replace the call to std::floor(d + .5) with a call to this function:

// this still has the same problems as the original rounding function
int round_up(double d)
    // return value will be coerced to int, and truncated as expected
    // you can then assign the int to a double, if desired
    return d + 0.5;

I think that suggests the following improvement:

// this won't work for negative d ...
// this may still round some numbers up when they should be rounded down
int round_up(double d)
    double floor;
    d = std::modf(d, &floor);
    return floor + (d + .5 + std::numeric_limits<double>::epsilon());

A simple note: std::numeric_limits<T>::epsilon() is defined as "the smallest number added to 1 that creates a number not equal to 1." You usually need to use a relative epsilon (i.e., scale epsilon somehow to account for the fact that you're working with numbers other than "1"). The sum of d, .5 and std::numeric_limits<double>::epsilon() should be near 1, so grouping that addition means that std::numeric_limits<double>::epsilon() will be about the right size for what we're doing. If anything, std::numeric_limits<double>::epsilon() will be too large (when the sum of all three is less than one) and may cause us to round some numbers up when we shouldn't.

Nowadays, you should consider std::nearbyint().

The accepted answer is correct if you are compiling to an x86 target that doesn't include SSE2. All modern x86 processors support SSE2, so if you can take advantage of it, you should:

-mfpmath=sse -msse2 -ffp-contract=off

Let's break this down.

-mfpmath=sse -msse2. This rounds inside SSE2 registers, which is much faster than storing every intermediate result to memory. Note that this is already the default on GCC for x86-64. From the GCC wiki:

On more modern x86 processors that support SSE2, specifying the compiler options -mfpmath=sse -msse2 ensures all float and double operations are performed in SSE registers and correctly rounded. These options do not affect the ABI and should therefore be used whenever possible for predictable numerical results.

-ffp-contract=off. Controlling rounding isn't enough for an exact match, however. FMA (fused multiply-add) instructions can change the rounding behavior versus its non-fused counterparts, so we need to disable it. This is the default on Clang, not GCC. As explained by this answer:

An FMA has only one rounding (it effectively keeps infinite precision for the internal temporary multiply result), while an ADD + MUL has two.

By disabling FMA, we get results that exactly match on debug and release, at the cost of some performance (and accuracy). We can still take advantage of other performance benefits of SSE and AVX.