Optimisations and undefined behaviour

[Richard Earnshaw]

On 09/11/15 14:29, Andrew Haley wrote:
> On 11/09/2015 02:16 PM, Florian Weimer wrote:
>> On 11/09/2015 03:08 PM, Andrew Haley wrote:
>>> On 11/09/2015 12:09 PM, Florian Weimer wrote:
>>>> On 11/09/2015 11:11 AM, Andrew Haley wrote:
>>>>> On 08/11/15 19:34, Segher Boessenkool wrote:
>>>>>> The compiler is free to transform it to
>>>>>>
>>>>>> int foo(int x) {
>>>>>> 	int t = x*x*x;
>>>>>> 	if (x > 1290) {
>>>>>> 		printf("X is wrong here %d, but we don't care\n", x);
>>>>>> 	}
>>>>>> 	return t;
>>>>>> }
>>>>>>
>>>>>> because x*x*x does not have any observable behaviour, and then it is
>>>>>> obvious it _can_ remove the printf and conditional.
>>>>
>>>> I'm not sure if this is a valid transformation for printf, even if
>>>> targets stdout and does not use any custom format specifiers.  Isn't it
>>>> a cancellation point?  But let's assume it's not.
>>>>
>>>>> Yes, that is correct.  And, indeed, the hardware is free to do taht
>>>>> too.  With speculative execution, the "as if" rule is not limited to
>>>>> the compiler.
>>>>
>>>> Can we disallow that optimization as a quality-of-implementation matter?
>>>>  What would be the benefit of such optimizations, other than
>>>> discouraging programmers from using C or C++?
>>>
>>> There isn't really any way to distinguish between wanted optimizations
>>> and unwanted ones.
>>
>> Of course there is—you define the semantics you want, and then any
>> optimization which breaks them is a bug.
>>
>>> If GCC determines that a statement is unreachable
>>> it can be deleted, and this depends on its knowledge of UB.  Like this:
>>>
>>> void foo(int b) {
>>>   if (b > 0) {
>>>     int m = b * 3 / 6;
>>>        if (m < 0)
>>>          die();
>>>      }
>>>    }
>>> }
>>>
>>> Deleting such unreachable code happens all the time.  IMO we should
>>> not disable this optimization.
>>
>> This is very different from the printf example.  The call to die is
>> unreachable according to the standard semantics.  The original printf
>> call is reachable, and according to my interpretation, the
>> transformation shown above is invalid because the abstract machine
>> performs the side effect from the printf before undefined behavior is
>> reached.
> 
> Here it is again:
> 
> int foo(int x) {
> 	if (x > 1290) {
> 		printf("X is wrong here %d, but we don't care\n", x);
> 	}
> 	return x*x*x;
> 
> Here, the printf writes to a stream then the UB happens.  


Not if setvbuf has been used to make the stream unbuffered.

> But the
> stream is buffered and the UB kills the process before the stream is
> flushed.  There is nothing in the C specification to prevent this, and
> neither should there be.  I don't think it's even possible.
> 
>>>> I'm worried that this particular line of argument would also allow the
>>>> movement of undefined behavior which occurs after an infinite loop in
>>>> front of it, even if this loop performs I/O.
>>>
>>> Sure.  But it can already do that even if the compiler does not move
>>> anything.  The I/O writes to a stream, the UB causes a segfault which
>>> kills a process, the stream never gets written.
>>
>> Based on my C semantics, the UB is never reached because the loop never
>> exits.
> 
> a.  You have your own C semantics?
> b.  Which loop?  You need to let us look at it.
> 
> I don't think that there is a program which will exhibit such
> behaviour.
> 
>> I just think that C semantics which only deal with terminating programs,
>> Turing-machine-style, are not very useful for the programs we generally
>> write.
> 
> This is an incomprehensible statement.  At leas, I don't know what it
> means.
> 
> Andrew.
>