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