Comparing D vs C++ (wierd behaviour of C++)

[Patrick Schluter]

On Tuesday, 24 July 2018 at 20:59:22 UTC, Patrick Schluter wrote:
> On Tuesday, 24 July 2018 at 19:24:05 UTC, Ecstatic Coder wrote:
>> On Tuesday, 24 July 2018 at 15:08:35 UTC, Patrick Schluter 
>> wrote:
>>> On Tuesday, 24 July 2018 at 14:41:17 UTC, Ecstatic Coder 
>>> wrote:
>>>> On Tuesday, 24 July 2018 at 14:08:26 UTC, Daniel Kozak wrote:
>>>>> [...]
>>>>
>>>> As the C++ char are signed by default, when you accumulate 
>>>> several shifted 8 bit -1 into a char result and then store 
>>>> it in a 64 bit unsigned buffer, you get -1 in 64 bits : 
>>>> 18446744073709551615.
>>>
>>> That's not exactly what happens here. There's no 64 bit 
>>> buffer.
>>
>> Sure about that ? ;)
>
> Yes, there are no "buffers" only register and a place on the 
> stack for the variable i.
>
> As said it's undefined behaviour so anything goes. I just 
> checked on godbolt what code is generated. 
> https://godbolt.org/g/wxqfmM
> So with -O0 this happens:
> From line 41 to line 77 the instruction to make the 
> calculation. At line 78
> mov DWORD PTR [rbp-40], eax which is writing out 32 bits to 
> reserved space of i.
> At line 85  mov eax, DWORD PTR [rbp-40] reloads that value in 
> eax, this annuls the high part of RAX => RAX contains 
> 0x0000_0000_FFFF_FFFF

what I forgot to mention, for the compiler the type deduction for 
the >> operator is done with the i variable, so it chooses the 
right template with unsigned int. For the optimized code as the 
calculation is done during compilation and there is no spill to 
the variable the type deduction for the >> operator for cout is 
done with that internal promoted temporary value and it deduces 
it as long (funnily declaring i as volatile doesn't change that 
even if the value is spilled to the stack).

>
> On the -O2 version it's even simpler. The calculation is done 
> at compile time and the endresult -1 is put directly to the 
> output. The test is even removed. Everything happens in the 
> compiler.