Re: Returning a reference from a function.
On Jul 12, 4:42 pm, Lionel B <m...@privacy.net> wrote:
On Thu, 12 Jul 2007 17:09:47 +0300, Juha Nieminen wrote:
Lionel B wrote:
Of course it's also *possible* (i.e. permitted by the language) - but
quite likely disastrous! - if the returned reference *is* to a local
$ g++ -std=c++98 -pedantic scratch.cpp scratch.cpp: In function ?int&
foo()?: scratch.cpp:3: warning: reference to local variable ?a?
(note warning, but no error).
Why is it not an error?
Because the Standard says it's not. I can't think off the top
of my head a valid scenario for doing this (no doubt someone
here can...), but there you go... C++ is renowned for
supplying plenty of rope to hang yourself.
The reason the standard says it doesn't require a diagnostic is
because it's impossible to determine in all cases. The standard
doesn't have too many options: if it's an error (it is), then it
is either undefined behavior, or a diagnostic is required. And
since it is impossible for a compiler to generate the diagnostic
in every case...
James Kanze (Gabi Software) email: firstname.lastname@example.org
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