Optimisations and undefined behaviour

[Richard Earnshaw]

On 09/11/15 14:56, Richard Earnshaw wrote:
> 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.
Sorry, that shouldn't appear here...

> 
>> 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.

... it should be here.  If the stream is unbuffered, then the ordering
is certainly well defined.

R.


>>
>>>>> 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.
>>
>