Why is returning a reference to a function local value not a compile error?

Question

The following code invokes undefined behaviour.

int& foo()
{
  int bar = 1234;
  return bar;
}

g++ issues a warning:

Answer 1

For the same reason C allows you to return a pointer to a memory block that's been freed.

It's valid according to the language specification. It's a horribly bad idea (and is nowhere close to being guaranteed to work) but it's still valid inasmuch as it's not forbidden.

If you're asking why the standard allows this, it's probably because, when references were introduced, that's the way they worked. Each iteration of the standard has certain guidelines to follow (such as minimising the possibility of "breaking changes", those that render existing well-formed programs invalid) and the standard is an agreement between user and implementer, with undoubtedly more implementers than users sitting on the committees :-)

It may be worth pushing that idea through as a potential change and seeing what ISO say but I suspect it would be considered one of those "breaking changes" and therefore very suspect.