Re: Temporary objects and operators overloading
Francis Glassborow wrote:
James Hopkin wrote:
Francis Glassborow wrote:
SG wrote:
I actually tested it a couple of weeks ago with G++ 4.3.3:
Foo operator+(Foo const & lhs, Foo const & rhs){
return Foo(lhs) += rhs; // No RVO is done
}
You mean it called the copy ctor twice? I am surprised.
Looking at 12.8/15 (the RVO):
when a temporary class object that has not been bound to a
reference (12.2) would be copied to a class object with the
same cv-unqualified type, the copy operation can be omitted
by constructing the temporary object directly into the
target of the omitted copy
it explicitly rules out eliding the copy when there's a reference
involved. The compiler *has* to call the copy constructor twice
(modulo the 'as if' rule).
But Foo(lhs) += rhs;
makes no attempt to bind the temporary being used to a reference.
Does "return *this;" in operator+= count as creating a reference?
Anyhow, I think I explained it well from a practical point of view.
There is no difference between the following two:
return Foo(lhs) += rhs;
return Foo(lhs).operator+=(rhs);
where operator+= is a function that returns a reference.
Unfortunately, the function signature
Foo& Foo::operator+=(Foo const&);
doesn't promise that the function returns *this. Therefore, the
compiler doesn't know in advance to what object the return value will
refer to. Because of that, it has to copy construct the return value
from whatever operator+= returns and allocate additional space for the
temporary object Foo(lhs).
In case operator+= is inlined the compiler could do better. Whether
that elision is legal is a question I cannot answer.
Cheers!
SG
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