Re: Strange warning from g++ "returning reference to temporary"

From:

Mateusz Adamczyk <matek_a@tlen.pl>

Newsgroups:

comp.lang.c++.moderated

Date:

Fri, 5 Dec 2008 20:48:21 CST

Message-ID:

<3692a878-1237-46ce-a470-3da837030985@l42g2000yqe.googlegroups.com>

On 5 Gru, 02:18, irotas <goo...@irotas.net> wrote:

Consider the following code:

struct Foo
{

operator const char* const&() const
{
return s; // <-- "warning: returning reference to temporary"
}

operator char*&()
{
return s; // <-- no warning!
}

char* s;

};

Aside from the "why would you want to do that anyway?" (there is a
reason!), could someone please explain the warning, and why there is
no warning on the non-const version?

Should there be a warning on the non-const version, or is it actually
OK?

Thanks!

-Adam

I suppose, that in the first function compiler creates temporary:
'const char* tmp = s' and later returns reference to it. That's reason
for warning. In the second function it's only returning reference to
s. That's why there is no warning.

Regards,
Mateusz

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