cannot call impure function ~this

[Namespace]

On Wednesday, 16 October 2013 at 07:32:27 UTC, monarch_dodra 
wrote:
> On Wednesday, 16 October 2013 at 07:27:25 UTC, Namespace wrote:
>> On Wednesday, 16 October 2013 at 07:23:45 UTC, monarch_dodra 
>> wrote:
>>> On Tuesday, 15 October 2013 at 21:37:40 UTC, Namespace wrote:
>>>> I get this error:
>>>> ----
>>>> /d701/f223.d(11): Error: pure function 'f223.getA' cannot 
>>>> call impure function 'f223.A.~this'
>>>> ----
>>>>
>>>> with this code:
>>>> ----
>>>> import std.stdio;
>>>>
>>>> struct A {
>>>> public:
>>>> 	~this() {
>>>> 		writeln("DTor");
>>>> 	}
>>>> }
>>>>
>>>> A getA() pure nothrow {
>>>> 	return A();
>>>> }
>>>>
>>>> void main()
>>>> {
>>>> 	A a = getA();
>>>> 	
>>>> 	writeln("end of main");
>>>> }
>>>> ----
>>>>
>>>> But without pure and nothrow I get this output:
>>>> ----
>>>> end of main
>>>> DTor
>>>> ----
>>>>
>>>> Why the compiler thinks that the function should/could call 
>>>> A::~this?
>>>
>>> It could have something to do with the fact that RVO is an 
>>> optimization *opportunity* that the compiler is allowed to go 
>>> for, even if it changes the program output.
>>>
>>> Hoewever, being an *opportunity*, the compiler still has to 
>>> make sure the code is valid without said optimization, which 
>>> in this case, isn't: getA would destroy it's temporary after 
>>> blitting it on the stac, leading to an impure call.
>>
>> So it _could_ be impure, but mostly it isn't, right?
>
> I guess that's one way to put it. I'd say it *is* impure, but 
> all its impure bits have been optimized out. That's my 
> explanation anyways.
>
> I'm curious: Is this a problem for you? The function calling 
> getA *can't* be pure either, so marking getA as pure is ...
>
> I was going to say useless, but I guess "pure" is always an 
> optimization opportunity for the compiler.
>
> I'd file an ER, you never know.
No, I was just curious what's behind it. So specifically why the 
compiler could call the destructor.