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

"Bo Persson" <>
Sat, 6 Dec 2008 15:00:28 CST
irotas 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?

The non-const version returns a reference to the actual type of the
member variable. That must be OK!

The const version returns a different type, and so possibly involves a
conversion - creating a temporary.

FWIW, the Comeau compiler gives the same warning, and it is usually
correct on language issues.

It is not at all obvious why anyone would return a reference to a
const pointer to a const object. :-)

Bo Persson

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