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I use the two following C++ compilers:
cl.exe : Microsoft (R) C/C++ Optimizing Compiler Version 19.00.24210 for x86
g++ : g++ (Ubuntu 5.2.1-22ubuntu2) 5.2.1 20151010
When using the built-in sine function, I get different results. This is not critical, but sometimes results are too significants for my use. Here is an example with a 'hard-coded' value:
printf("%f\n", sin(5451939907183506432.0));
Result with cl.exe:
0.528463
Result with g++:
0.522491
I know that g++'s result is more accurate and that I could use an additional library to get this same result, but that's not my point here. I would really understand what happens here: why is cl.exe that wrong?
Funny thing, if I apply a modulo of (2 * pi) on the param, then I get the same result than g++...
[EDIT] Just because my example looks crazy for some of you: this is a part of a pseudorandom number generator. It is not important to know if the result of the sine is accurate or not: we just need it to give some result.
You have a 19-digit literal, but double usually has 15-17 digit precision. As a result, you can get a small relative error (when converting to double), but big enough (in the context of sine calculation) absolute error.
Actually, different implementations of the standard library have differences in treating such large numbers. For example, in my environment, if we execute
std::cout << std::fixed << 5451939907183506432.0;
g++ result would be 5451939907183506432.000000
cl result would be 5451939907183506400.000000
The difference is because versions of cl earlier than 19 have a formatting algorithm that uses only a limited number of digits and fills the remaining decimal places with zero.
Furthermore, let's look at this code:
double a[1000];
for (int i = 0; i < 1000; ++i) {
a[i] = sin(5451939907183506432.0);
}
double d = sin(5451939907183506432.0);
cout << a[500] << endl;
cout << d << endl;
When executed with my x86 VC++ compiler the output is:
0.522491
0.528463
It appears that when filling the array sin is compiled to the call of __vdecl_sin2, and when there is a single operation, it is compiled to the call of __libm_sse2_sin_precise (with /fp:precise).
In my opinion, your number is too large for sin calculation to expect the same behavior from different compilers and to expect the correct behavior in general.
I think Sam's comment is closest to the mark. Whereas you're using a recentish version of GCC/glibc, which implements sin() in software (calculated at compile time for the literal in question), cl.exe for x86 likely uses the fsin instruction. The latter can be very imprecise, as described in the Random ASCII blog post, "Intel Underestimates Error Bounds by 1.3 quintillion".
Part of the problem with your example in particular is that Intel uses an imprecise approximation of pi when doing range reduction:
When doing range reduction from double-precision (53-bit mantissa) pi the results will have about 13 bits of precision (66 minus 53), for an error of up to 2^40 ULPs (53 minus 13).
According to cppreference:
The result may have little or no significance if the magnitude of arg is large
(until C++11)
It's possible that this is the cause of the problem, in which case you will want to manually do the modulo so that arg is not large.
Sample code:
#include <iostream>
#include <cmath>
#include <stdint.h>
using namespace std;
static bool my_isnan(double val) {
union { double f; uint64_t x; } u = { val };
return (u.x << 1) > (0x7ff0000000000000u << 1);
}
int main() {
cout << std::isinf(std::log(0.0)) << endl;
cout << std::isnan(std::sqrt(-1.0)) << endl;
cout << my_isnan(std::sqrt(-1.0)) << endl;
cout << __isnan(std::sqrt(-1.0)) << endl;
return 0;
}
Online compiler.
With -ffast-math, that code prints "0, 0, 1, 1" -- without, it prints "1, 1, 1, 1".
Is that correct? I thought that std::isinf/std::isnan should still work with -ffast-math in these cases.
Also, how can I check for infinity/NaN with -ffast-math? You can see the my_isnan doing this, and it actually works, but that solution is of course very architecture dependent. Also, why does my_isnan work here and std::isnan does not? What about __isnan and __isinf. Do they always work?
With -ffast-math, what is the result of std::sqrt(-1.0) and std::log(0.0). Does it become undefined, or should it be NaN / -Inf?
Related discussions: (GCC) [Bug libstdc++/50724] New: isnan broken by -ffinite-math-only in g++, (Mozilla) Bug 416287 - performance improvement opportunity with isNaN
Note that -ffast-math may make the compiler ignore/violate IEEE specifications, see http://gcc.gnu.org/onlinedocs/gcc-4.8.2/gcc/Optimize-Options.html#Optimize-Options :
This option is not turned on by any -O option besides -Ofast since it
can result in incorrect output for programs that depend on an exact
implementation of IEEE or ISO rules/specifications for math functions.
It may, however, yield faster code for programs that do not require
the guarantees of these specifications.
Thus, using -ffast-math you are not guaranteed to see infinity where you should.
In particular, -ffast-math turns on -ffinite-math-only, see http://gcc.gnu.org/wiki/FloatingPointMath which means (from http://gcc.gnu.org/onlinedocs/gcc-4.8.2/gcc/Optimize-Options.html#Optimize-Options )
[...] optimizations for floating-point arithmetic that assume that arguments and results are not NaNs or +-Infs
This means, by enabling the -ffast-math you make a promise to the compiler that your code will never use infinity or NaN, which in turn allows the compiler to optimize the code by, e.g., replacing any calls to isinf or isnan by the constant false (and further optimize from there). If you break your promise to the compiler, the compiler is not required to create correct programs.
Thus the answer quite simple, if your code may have infinities or NaN (which is strongly implied by the fact that you use isinf and isnan), you cannot enable -ffast-math as else you might get incorrect code.
Your implementation of my_isnan works (on some systems) because it directly checks the binary representation of the floating point number. Of course, the processor still might do (some) actual calculations (depending on which optimizations the compiler does), and thus actual NaNs might appear in memory and you can check their binary representation, but as explained above, std::isnan might have been replaced by the constant false. It might equally well happen that the compiler replaces, e.g., sqrt, by some version that doesn't even produce a NaN for input -1. In order to see which optimisations your compiler does, compile to assembler and look at that code.
To make a (not completely unrelated) analogy, if you're telling your compiler your code is in C++ you can not expect it to compile C code correctly and vice-versa (there are actual examples for this, e.g. Can code that is valid in both C and C++ produce different behavior when compiled in each language? ).
It is a bad idea to enable -ffast-math and use my_isnan because this will make everything very machine- and compiler-dependent you don't know what optimizations the compiler does overall, so there might be other hidden problems related to the fact that you are using non-finite maths but tell the compiler otherwise.
A simple fix is to use -ffast-math -fno-finite-math-only which would still give some optimizations.
It also might be that your code looks something like this:
filter out all infinities and NaNs
do some finite maths on the filtered values (by this I mean maths that is guaranteed to never create infinities or NaNs, this has to be very, very carefully checked)
In this case, you could split up your code and either use optimize #pragma or __attribute__ to turn -ffast-math (respectively -ffinite-math-only and -fno-finite-math-only) on and off selectively for the given pieces of code (however, I remember there being some trouble with some version of GCC related to this) or just split your code into separate files and compile them with different flags. Of course, this also works in more general settings if you can isolate the parts where infinities and NaNs might occur. If you can not isolate these parts, this is a strong indication that you can not use -ffinite-math-only for this code.
Finally, it's important to understand that -ffast-math is not a harmless optimization that simply makes your program faster. It does not only affect the performance of your code but also its correctness (and this on top of all the issues surrounding floating point numbers already, if I remember right William Kahan has a collection of horror stories on his homepage, see also What every programmer should know about floating point arithmetic). In short, you might get faster code, but also wrong or unexpected results (see below for an example). Hence, you should only use such optimizations when you really know what you are doing and you have made absolutely sure, that either
the optimizations don't affect the correctness of that particular code, or
the errors introduced by the optimization are not critical to the code.
Program code can actually behave quite differently depending on whether this optimization is used or not. In particular it can behave wrong (or at least very contrary to your expectations) when optimizations such as -ffast-math are enabled. Take the following program for example:
#include <iostream>
#include <limits>
int main() {
double d = 1.0;
double max = std::numeric_limits<double>::max();
d /= max;
d *= max;
std::cout << d << std::endl;
return 0;
}
will produce output 1 as expected when compiled without any optimization flag, but using -ffast-math, it will output 0.
I'm currently looking at code which does multi-precision floating-point arithmetic. To work correctly, that code requires values to be reduced to their final precision at well-defined points. So even if an intermediate result was computed to an 80 bit extended precision floating point register, at some point it has to be rounded to 64 bit double for subsequent operations.
The code uses a macro INEXACT to describe this requirement, but doesn't have a perfect definition. The gcc manual mentions -fexcess-precision=standard as a way to force well-defined precision for cast and assignment operations. However, it also writes:
‘-fexcess-precision=standard’ is not implemented for languages other than C
Now I'm thinking about porting those ideas to C++ (comments welcome if anyone knows an existing implementation). So it seems I can't use that switch for C++. But what is the g++ default behavior in absence of any switch? Are there more C++-like ways to control the handling of excess precision?
I guess that for my current use case, I'll probably use -mfpmath=sse in any case, which should not incur any excess precision as far as I know. But I'm still curious.
Are there more C++-like ways to control the handling of excess precision?
The C99 standard defines FLT_EVAL_METHOD, a compiler-set macro that defines how excess precision should happen in a C program (many C compilers still behave in a way that does not exactly conform to the most reasonable interpretation of the value of FP_EVAL_METHOD that they define: older GCC versions generating 387 code, Clang when generating 387 code, …). Subtle points in relation with the effects of FLT_EVAL_METHOD were clarified in the C11 standard.
Since the 2011 standard, C++ defers to C99 for the definition of FLT_EVAL_METHOD (header cfloat).
So GCC should simply allow -fexcess-precision=standard for C++, and hopefully it eventually will. The same semantics as that of C are already in the C++ standard, they only need to be implemented in C++ compilers.
I guess that for my current use case, I'll probably use -mfpmath=sse in any case, which should not incur any excess precision as far as I know.
That is the usual solution.
Be aware that C99 also defines FP_CONTRACT in math.h that you may want to look at: it relates to the same problem of some expressions being computed at a higher precision, striking from a completely different side (the modern fused-multiply-add instruction instead of the old 387 instruction set). This is a pragma to decide whether the compiler is allowed to replace source-level additions and multiplications with FMA instructions (this has the effect that the multiplication is virtually computed at infinite precision, because this is how this instruction works, instead of being rounded to the precision of the type as it would be with separate multiplication and addition instructions). This pragma has apparently not been incorporated in the C++ standard (as far as I can see).
The default value for this option is implementation-defined and some people argue for the default to be to allow FMA instructions to be generated (for C compilers that otherwise define FLT_EVAL_METHOD as 0).
You should, in C, future-proof
your code with:
#include <math.h>
#pragma STDC FP_CONTRACT off
And the equivalent incantation in C++ if your compiler documents one.
what is the g++ default behavior in absence of any switch?
I am afraid that the answer to this question is that GCC's behavior, say, when generating 387 code, is nonsensical. See the description of the situation that motivated Joseph Myers to fix the situation for C. If g++ does not implement -fexcess-precision=standard, it probably means that 80-bit computations are randomly rounded to the precision of the type when the compiler happened to have to spill some floating-point registers to memory, leading the program below to print "foo" in some circumstances outside the programmer's control:
if (x == 0.0) return;
... // code that does not modify x
if (x == 0.0) printf("foo\n");
… because the code in the ellipsis caused x, that was held in an 80-bit floating-point register, to be spilt to a 64-bit slot on the stack.
But what is the g++ default behavior in absence of any switch?
I found one answer myself via an experiment, using the following code:
#include <stdio.h>
#include <stdlib.h>
int main(int argc, char** argv) {
double a = atof("1.2345678");
double b = a*a;
printf("%.20e\n", b - 1.52415765279683990130);
return 0;
}
If b is rounded (-fexcess-precision=standard), then the result is zero. Otherwise (-fexcess-precision=fast) it is something like 8e-17. Compiling with -mfpmath=387 -O3, I could reproduce both cases for gcc-4.8.2. For g++-4.8.2 I get an error for -fexcess-precision=standard if I try that, and without a flag I get the same behavior as -fexcess-precision=fast gives for C. Adding -std=c++11 does not help. So now the suspicion already voiced by Pascal is official: g++ does not necessarily round everywhere it should.
My questions are divided into three parts
Question 1
Consider the below code,
#include <iostream>
using namespace std;
int main( int argc, char *argv[])
{
const int v = 50;
int i = 0X7FFFFFFF;
cout<<(i + v)<<endl;
if ( i + v < i )
{
cout<<"Number is negative"<<endl;
}
else
{
cout<<"Number is positive"<<endl;
}
return 0;
}
No specific compiler optimisation options are used or the O's flag is used. It is basic compilation command g++ -o test main.cpp is used to form the executable.
The seemingly very simple code, has odd behaviour in SUSE 64 bit OS, gcc version 4.1.2. The expected output is "Number is negative", instead only in SUSE 64 bit OS, the output would be "Number is positive".
After some amount of analysis and doing a 'disass' of the code, I find that the compiler optimises in the below format -
Since i is same on both sides of comparison, it cannot be changed in the same expression, remove 'i' from the equation.
Now, the comparison leads to if ( v < 0 ), where v is a constant positive, So during compilation itself, the else part cout function address is added to the register. No cmp/jmp instructions can be found.
I see that the behaviour is only in gcc 4.1.2 SUSE 10. When tried in AIX 5.1/5.3 and HP IA64, the result is as expected.
Is the above optimisation valid?
Or, is using the overflow mechanism for int not a valid use case?
Question 2
Now when I change the conditional statement from if (i + v < i) to if ( (i + v) < i ) even then, the behaviour is same, this atleast I would personally disagree, since additional braces are provided, I expect the compiler to create a temporary built-in type variable and them compare, thus nullify the optimisation.
Question 3
Suppose I have a huge code base, an I migrate my compiler version, such bug/optimisation can cause havoc in my system behaviour. Ofcourse from business perspective, it is very ineffective to test all lines of code again just because of compiler upgradation.
I think for all practical purpose, these kinds of error are very difficult to catch (during upgradation) and invariably will be leaked to production site.
Can anyone suggest any possible way to ensure to ensure that these kind of bug/optimization does not have any impact on my existing system/code base?
PS :
When the const for v is removed from the code, then optimization is not done by the compiler.
I believe, it is perfectly fine to use overflow mechanism to find if the variable is from MAX - 50 value (in my case).
Update(1)
What would I want to achieve? variable i would be a counter (kind of syncID). If I do offline operation (50 operation) then during startup, I would like to reset my counter, For this I am checking the boundary value (to reset it) rather than adding it blindly.
I am not sure if I am relying on the hardware implementation. I know that 0X7FFFFFFF is the max positive value. All I am doing is, by adding value to this, I am expecting the return value to be negative. I don't think this logic has anything to do with hardware implementation.
Anyways, all thanks for your input.
Update(2)
Most of the inpit states that I am relying on the lower level behavior on overflow checking. I have one questions regarding the same,
If that is the case, For an unsigned int how do I validate and reset the value during underflow or overflow? like if v=10, i=0X7FFFFFFE, I want reset i = 9. Similarly for underflow?
I would not be able to do that unless I check for negativity of the number. So my claim is that int must return a negative number when a value is added to the +MAX_INT.
Please let me know your inputs.
It's a known problem, and I don't think it's considered a bug in the compiler. When I compile with gcc 4.5 with -Wall -O2 it warns
warning: assuming signed overflow does not occur when assuming that (X + c) < X is always false
Although your code does overflow.
You can pass the -fno-strict-overflow flag to turn that particular optimization off.
Your code produces undefined behavior. C and C++ languages has no "overflow mechanism" for signed integer arithmetic. Your calculations overflow signed integers - the behavior is immediately undefined. Considering it form "a bug in the compiler or not" position is no different that attempting to analyze the i = i++ + ++i examples.
GCC compiler has an optimization based on that part of the specification of C/C++ languages. It is called "strict overflow semantics" or something lake that. It is based on the fact that adding a positive value to a signed integer in C++ always produces a larger value or results in undefined behavior. This immediately means that the compiler is perfectly free to assume that the sum is always larger. The general nature of that optimization is very similar to the "strict aliasing" optimizations also present in GCC. They both resulted in some complaints from the more "hackerish" parts of GCC user community, many of whom didn't even suspect that the tricks they were relying on in their C/C++ programs were simply illegal hacks.
Q1: Perhaps, the number is indeed positive in a 64bit implementation? Who knows? Before debugging the code I'd just printf("%d", i+v);
Q2: The parentheses are only there to tell the compiler how to parse an expression. This is usually done in the form of a tree, so the optimizer does not see any parentheses at all. And it is free to transform the expression.
Q3: That's why, as c/c++ programmer, you must not write code that assumes particular properties of the underlying hardware, such as, for example, that an int is a 32 bit quantity in two's complement form.
What does the line:
cout<<(i + v)<<endl;
Output in the SUSE example? You're sure you don't have 64bit ints?
OK, so this was almost six years ago and the question is answered. Still I feel that there are some bits that have not been adressed to my satisfaction, so I add a few comments, hopefully for the good of future readers of this discussion. (Such as myself when I got a search hit for it.)
The OP specified using gcc 4.1.2 without any special flags. I assume the absence of the -O flag is equivalent to -O0. With no optimization requested, why did gcc optimize away code in the reported way? That does seem to me like a compiler bug. I also assume this has been fixed in later versions (for example, one answer mentions gcc 4.5 and the -fno-strict-overflow optimization flag). The current gcc man page states that -fstrict-overflow is included with -O2 or more.
In current versions of gcc, there is an option -fwrapv that enables you to use the sort of code that caused trouble for the OP. Provided of course that you make sure you know the bit sizes of your integer types. From gcc man page:
-fstrict-overflow
.....
See also the -fwrapv option. Using -fwrapv means that integer signed overflow
is fully defined: it wraps. ... With -fwrapv certain types of overflow are
permitted. For example, if the compiler gets an overflow when doing arithmetic
on constants, the overflowed value can still be used with -fwrapv, but not otherwise.
I'm writing a C++ program that doesn't work (I get a segmentation fault) when I compile it with optimizations (options -O1, -O2, -O3, etc.), but it works just fine when I compile it without optimizations.
Is there any chance that the error is in my code? or should I assume that this is a bug in GCC?
My GCC version is 3.4.6.
Is there any known workaround for this kind of problem?
There is a big difference in speed between the optimized and unoptimized version of my program, so I really need to use optimizations.
This is my original functor. The one that works fine with no levels of optimizations and throws a segmentation fault with any level of optimization:
struct distanceToPointSort{
indexedDocument* point ;
distanceToPointSort(indexedDocument* p): point(p) {}
bool operator() (indexedDocument* p1,indexedDocument* p2){
return distance(point,p1) < distance(point,p2) ;
}
} ;
And this one works flawlessly with any level of optimization:
struct distanceToPointSort{
indexedDocument* point ;
distanceToPointSort(indexedDocument* p): point(p) {}
bool operator() (indexedDocument* p1,indexedDocument* p2){
float d1=distance(point,p1) ;
float d2=distance(point,p2) ;
std::cout << "" ; //without this line, I get a segmentation fault anyways
return d1 < d2 ;
}
} ;
Unfortunately, this problem is hard to reproduce because it happens with some specific values. I get the segmentation fault upon sorting just one out of more than a thousand vectors, so it really depends on the specific combination of values each vector has.
Now that you posted the code fragment and a working workaround was found (#Windows programmer's answer), I can say that perhaps what you are looking for is -ffloat-store.
-ffloat-store
Do not store floating point variables in registers, and inhibit other options that might change whether a floating point value is taken from a register or memory.
This option prevents undesirable excess precision on machines such as the 68000 where the floating registers (of the 68881) keep more precision than a double is supposed to have. Similarly for the x86 architecture. For most programs, the excess precision does only good, but a few programs rely on the precise definition of IEEE floating point. Use -ffloat-store for such programs, after modifying them to store all pertinent intermediate computations into variables.
Source: http://gcc.gnu.org/onlinedocs/gcc-3.4.6/gcc/Optimize-Options.html
I would assume your code is wrong first.
Though it is hard to tell.
Does your code compile with 0 warnings?
g++ -Wall -Wextra -pedantic -ansi
Here's some code that seems to work, until you hit -O3...
#include <stdio.h>
int main()
{
int i = 0, j = 1, k = 2;
printf("%d %d %d\n", *(&j-1), *(&j), *(&j+1));
return 0;
}
Without optimisations, I get "2 1 0"; with optimisations I get "40 1 2293680". Why? Because i and k got optimised out!
But I was taking the address of j and going out of the memory region allocated to j. That's not allowed by the standard. It's most likely that your problem is caused by a similar deviation from the standard.
I find valgrind is often helpful at times like these.
EDIT: Some commenters are under the impression that the standard allows arbitrary pointer arithmetic. It does not. Remember that some architectures have funny addressing schemes, alignment may be important, and you may get problems if you overflow certain registers!
The words of the [draft] standard, on adding/subtracting an integer to/from a pointer (emphasis added):
"If both the pointer operand and the result point to elements of the same array object, or one past the last element of the array object, the evaluation shall not produce an overflow; otherwise, the behavior is undefined."
Seeing as &j doesn't even point to an array object, &j-1 and &j+1 can hardly point to part of the same array object. So simply evaluating &j+1 (let alone dereferencing it) is undefined behaviour.
On x86 we can be pretty confident that adding one to a pointer is fairly safe and just takes us to the next memory location. In the code above, the problem occurs when we make assumptions about what that memory contains, which of course the standard doesn't go near.
As an experiment, try to see if this will force the compiler to round everything consistently.
volatile float d1=distance(point,p1) ;
volatile float d2=distance(point,p2) ;
return d1 < d2 ;
The error is in your code. It's likely you're doing something that invokes undefined behavior according to the C standard which just happens to work with no optimizations, but when GCC makes certain assumptions for performing its optimizations, the code breaks when those assumptions aren't true. Make sure to compile with the -Wall option, and the -Wextra might also be a good idea, and see if you get any warnings. You could also try -ansi or -pedantic, but those are likely to result in false positives.
You may be running into an aliasing problem (or it could be a million other things). Look up the -fstrict-aliasing option.
This kind of question is impossible to answer properly without more information.
It is very seldom the compiler fault, but compiler do have bugs in them, and them often manifest themselves at different optimization levels (if there is a bug in an optimization pass, for example).
In general when reporting programming problems: provide a minimal code sample to demonstrate the issue, such that people can just save the code to a file, compile and run it. Make it as easy as possible to reproduce your problem.
Also, try different versions of GCC (compiling your own GCC is very easy, especially on Linux). If possible, try with another compiler. Intel C has a compiler which is more or less GCC compatible (and free for non-commercial use, I think). This will help pinpointing the problem.
It's almost (almost) never the compiler.
First, make sure you're compiling warning-free, with -Wall.
If that didn't give you a "eureka" moment, attach a debugger to the least optimized version of your executable that crashes and see what it's doing and where it goes.
5 will get you 10 that you've fixed the problem by this point.
Ran into the same problem a few days ago, in my case it was aliasing. And GCC does it differently, but not wrongly, when compared to other compilers. GCC has become what some might call a rules-lawyer of the C++ standard, and their implementation is correct, but you also have to be really correct in you C++, or it'll over optimize somethings, which is a pain. But you get speed, so can't complain.
I expect to get some downvotes here after reading some of the comments, but in the console game programming world, it's rather common knowledge that the higher optimization levels can sometimes generate incorrect code in weird edge cases. It might very well be that edge cases can be fixed with subtle changes to the code, though.
Alright...
This is one of the weirdest problems I've ever had.
I dont think I have enough proof to state it's a GCC bug, but honestly... It really looks like one.
This is my original functor. The one that works fine with no levels of optimizations and throws a segmentation fault with any level of optimization:
struct distanceToPointSort{
indexedDocument* point ;
distanceToPointSort(indexedDocument* p): point(p) {}
bool operator() (indexedDocument* p1,indexedDocument* p2){
return distance(point,p1) < distance(point,p2) ;
}
} ;
And this one works flawlessly with any level of optimization:
struct distanceToPointSort{
indexedDocument* point ;
distanceToPointSort(indexedDocument* p): point(p) {}
bool operator() (indexedDocument* p1,indexedDocument* p2){
float d1=distance(point,p1) ;
float d2=distance(point,p2) ;
std::cout << "" ; //without this line, I get a segmentation fault anyways
return d1 < d2 ;
}
} ;
Unfortunately, this problem is hard to reproduce because it happens with some specific values. I get the segmentation fault upon sorting just one out of more than a thousand vectors, so it really depends on the specific combination of values each vector has.
Wow, I didn't expect answers so quicly, and so many...
The error occurs upon sorting a std::vector of pointers using std::sort()
I provide the strict-weak-ordering functor.
But I know the functor I provide is correct because I've used it a lot and it works fine.
Plus, the error cannot be some invalid pointer in the vector becasue the error occurs just when I sort the vector. If I iterate through the vector without applying std::sort first, the program works fine.
I just used GDB to try to find out what's going on. The error occurs when std::sort invoke my functor. Aparently std::sort is passing an invalid pointer to my functor. (of course this happens with the optimized version only, any level of optimization -O, -O2, -O3)
as other have pointed out, probably strict aliasing.
turn it of in o3 and try again. My guess is that you are doing some pointer tricks in your functor (fast float as int compare? object type in lower 2 bits?) that fail across inlining template functions.
warnings do not help to catch this case. "if the compiler could detect all strict aliasing problems it could just as well avoid them" just changing an unrelated line of code may make the problem appear or go away as it changes register allocation.
As the updated question will show ;) , the problem exists with a std::vector<T*>. One common error with vectors is reserve()ing what should have been resize()d. As a result, you'd be writing outside array bounds. An optimizer may discard those writes.
post the code in distance! it probably does some pointer magic, see my previous post. doing an intermediate assignment just hides the bug in your code by changing register allocation. even more telling of this is the output changing things!
The true answer is hidden somewhere inside all the comments in this thread. First of all: it is not a bug in the compiler.
The problem has to do with floating point precision. distanceToPointSort should be a function that should never return true for both the arguments (a,b) and (b,a), but that is exactly what can happen when the compiler decides to use higher precision for some data paths. The problem is especially likely on, but by no means limited to, x86 without -mfpmath=sse. If the comparator behaves that way, the sort function can become confused, and the segmentation fault is not surprising.
I consider -ffloat-store the best solution here (already suggested by CesarB).