Stupid scope destruction question

[monarch_dodra]

On Monday, 24 September 2012 at 07:25:28 UTC, Denis Shelomovskij 
wrote:
> 20.09.2012 15:35, monarch_dodra пишет:
>>
>> AFAIK, if the rules are the same in C++ (which they probably 
>> are), then:
>> "Any object constructed during argument passing will remain 
>> valid for
>> the duration of the call. It will go out of scope once the 
>> function has
>> finished returning, and after the return value has itself gone 
>> out of
>> scope and been destroyed."
>>
>
> Thanks, looks like D does have C++ behaviour here. But your 
> last statement about return value is incorrect. More than that 
> function call doesn't change anything.
>
> Correct answers are here:
> * `12.2 Temporary objects [class.temporary]` section of C++ 
> standard
> * 
> http://stackoverflow.com/questions/2506793/c-life-span-of-temporary-arguments
> * 
> http://stackoverflow.com/questions/5459759/full-expression-boundaries-and-lifetime-of-temporaries

How is my statement incorrect? The "function call" itself doesn't 
change anything sure, since it is more generally a "full 
expression": The return value itself is created *during* that 
full expression, but after the creation of the arguments. Last 
in, first out, it is destroyed before the passed in arguments.

Unless you were playing on the words "gone out of scope" (which 
is indeed not the 100% correct), I don't see how my statement is 
incorrect.