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

Alex Shulgin <>
Fri, 5 Dec 2008 20:34:30 CST
On Dec 5, 3:18 am, 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

Because `const char*' is a pointer to const char, and Foo::s is a
pointer to non-const char: compiler automatically casts pointer to non-
const to const, so here's a temporary. The following is equivalent of
code generated by a typical compiler:

     operator const char* const&() const
       const char *tmp = s;
       return tmp;

With non-const vesion types already match, so there's no need in
conversion, hence no temporary is created.

You might want to remove reference symbol from operator prototype.


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