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. 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.
