[LLVMdev] interesting possible compiler bug

Richard Smith richard at metafoo.co.uk
Tue Oct 2 15:01:49 PDT 2012

See also this proposal for the next C++ committee meeting, which aims to
clarify this case and others like it:


On Tue, Oct 2, 2012 at 9:49 AM, Reed Kotler <rkotler at mips.com> wrote:

> I've sent this issue to some friends on the C++ committee.
> The first response I received indicates that the person thinks the
> optimizer is within it's rights to optimize away the call to malloc.
> Here are some points, extracted from the response:
> "There is no observable behavior (a technical term in the standard) that
> violates requirements of the standard. In particular, malloc in the
> abstract machine defined by the standard need not ever fail."
> "On linux in particular, malloc (almost) never fails, because Linux does
> not actually make malloc'd memory available until it is used. Here the
> allocated memory is never used, so if the compiler recognizes malloc as a
> standard library function with well-defined semantics, it can eliminate the
> actual call and pretend it succeeded.
> Since variable curr is not visible outside main and is not declared
> volatile, it can be optimized away."
> On 10/02/2012 02:12 AM, Nick Lewycky wrote:
>> David Chisnall wrote:
>>> On 2 Oct 2012, at 03:40, Nick Lewycky wrote:
>>>  As far as I know, this optimization is legal. Fix the test with a
>>>> volatile pointer:
>>> Why would that be required?
>> It isn't. My suggestion for fixing the test is to make use of the
>> returned pointer in a fashion that the compiler is forbidden to elide
>> it: make it an observable side-effect using volatile. Another way would
>> be to take the pointer and print it to file or stdout. Perhaps you can
>> think of others.
>>    malloc() is defined by the standard to return a pointer that is
>> distinct from any other valid pointer, or NULL.  Any optimisation that
>> makes any assumptions about its next value is invalid.
>> Nowhere do we make assumptions about malloc's next value.
>> This is a straight-forward application of the as-if rule. The malloc
>> call behaves as-if it allocated memory. Because we prove that the code
>> doesn't use that memory, we can get away with allocating no memory at
>> all and not change the behaviour of the program.
>> But we did change the behaviour of the program, didn't we? Well, we
>> haven't changed the behaviour of the program in any way that is
>> observable through a well-defined mechanism. Crashes, running out of
>> memory, or asking the kernel for your processes' memory usage isn't
>> behaviour you get to rely on. The first two really aren't defined in the
>> language, and the last one goes through I/O which is permitted to do its
>> own thing. (For instance, we don't forbid constant-folding "1+1" because
>> a program may fopen and disassemble itself to look for the "add 1, 1"
>> instruction.)
>>  int main() {
>>>>    volatile char *curr;
>>>>    do {
>>>>      curr = malloc(1);
>>>>      int i = *curr;
>>> This, in particular, looks very wrong.  If curr is void, then you are
>>> dereferencing an invalid pointer, and so you are doing something
>>> undefined.
>> Do you mean, if curr is NULL? It's a char*, not void*.
>> In fact, this version of the code is completely free to elide the
>> conditional loop, because by dereferencing the pointer you are asserting
>> that it is not NULL (or, at least, that if it is then after this point
>> the program is in an undefined state and so any behaviour is legal) and
>> so it is completely free to generate the code that it in fact does
>> generate without this test.  So here we have another bug, because the
>> testq in your output is redundant after the movb.
>> Yes, good point, I totally missed the NULL dereference. I haven't
>> checked what happens if you write:
>>    curr = malloc(1);
>>    if (curr)
>>      int i = *curr;
>> but I expect that would work.
>> Nick
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