function not callable using struct constructor

[alex burton]

On Friday, 3 January 2014 at 03:59:50 UTC, Jonathan M Davis wrote:
> On Friday, January 03, 2014 03:13:12 alex burton wrote:
>> struct Foo
>> {
>> };
>> 
>> void bar(ref Foo f)
>> {
>> }
>> 
>> void main()
>> {
>> bar(Foo()); //Error: function test.bar (ref Foo f) is not
>> callable using argument types (Foo)
>> }
>> 
>> I get the above error with 2.064 not with 2.060.
>> Is it a bug ?
>> 
>> Is it a feature ?
>> If so :
>> Why can't I take a non const ref to a temp struct - It might 
>> look
>> a bit silly but I might be doing it to avoid copying.
>> 
>> I could still do this :
>> 
>> void main()
>> {
>> Foo f = Foo();
>> bar(f);
>> }
>> 
>> Which is equivalent AFAIK
>
> ref parameters only accept lvalues. Foo() creates a temporary 
> and is not an
> lvalue. If you want a function to accept rvalues, it has to not 
> take its
> argument by ref. If you want to avoid copying lvalues and still 
> accept
> rvalues, then you need to overload the function (one which 
> takes ref and one
> which doesn't). And unlike in C++, constness has no effect on 
> whether ref
> accepts rvalues. ref only ever accepts lvalues.
>
> However, in the case of rvalues, ref doesn't save you anything 
> performance-
> wise. No copy will occur. Rather, because it's an rvalue, it 
> can just move it
> into the function's parameter rather than copying it. ref only 
> saves you
> copying with lvalues.
>
> - Jonathan M Davis

Thanks for this, although I tested in 2.064 and the above code 
works using void bar(in Foo f), which would seem to contracdict 
"constness has no effect on whether ref accepts rvalues" if I 
understand it correctly.

After looking at the asm, it appears that the compiler doesn't 
copy the struct when using a void bar(Foo f), but also can avoid 
copying if a second function like bar is called from bar.

It appears that passing by value is what I need, and I should 
trust the compiler.