Re: Strange warning from g++ "returning reference to temporary"
Bo Persson wrote:
blargg wrote:
Bo Persson wrote:
The const version returns a different type, and so possibly
involves a conversion - creating a temporary.
[...]
I'm confused, since as far as I understand, returning a pointer
rather than reference WOULD work (4.4 para 4), without creating any
temporaries:
struct Foo
{
char* s;
operator const char* const*() const { return &s; } // OK, right?
operator const char* const&() const { return s; } // why not OK?
};
If the reference version really is invalid (and the pointer version
not), then this is one glaring difference between references and
pointers where one would expect them to behave the same.
Well, references and pointers are different. :-)
I should have stated that this is not a beginner question, thus it
warrants close attention to details. :)
In the pointer case, you actually return the pointer BY VALUE. That
way you get a temporary anyway, but it is copied, so it doesn't
matter.
The reference version returns a reference to a temporary that
immediately goes out of scope. That's serious!
Let me add a third function that illustrates my point:
struct Foo
{
char* s;
operator const char* const*() const { return &s; } // OK
operator const char* const&() const { return s; } // not OK
// you're saying this is what the above does:
operator const char* const&() const
{
const char* const p = s;
return p; // NOT OK, returning reference to local
}
// I'm saying, why doesn't it instead do this:
operator const char* const&() const
{
const char* const* p = &s; // OK
return *p; // OK, reference to original s object
}
};
The last operator function shows that a temporary isn't necessary in
this
case (just as it's not necessary in the pointer case).
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