Re: return reference to local variable, good code?
On Dec 17, 11:56 am, Pavel Shved <Pavel.Sh...@gmail.com> wrote:
On Dec 16, 13:39, James Kanze <james.ka...@gmail.com> wrote:
On Dec 16, 7:11 am, "Alf P. Steinbach" <al...@start.no> wrote:
* yanlinlin:
On 12=E6=9C=8816=E6=97=A5, =E4=B8=8B=E5=8D=881=E6=97=B649=E5=88=86, =
George2 <george4acade...@yahoo.com> wrote:
I am reading some code from other people, there are some code like
this,
[Code]
class Foo {
};
Foo& func()
{
Foo foo;
return foo;
}
int main()
{
Foo& foo = func();
return 0;}
[/Code]
I want to confirm with you that it is not good code, since
we return a reference to local object instance, right? Even
if the code has expected function currently.
I'm afraid this code can not be compiled. To return a
reference of local non-static object is not allowed.
It's allowed and it should compile.
A compiler that doesn't allow it is non-conforming.
However, using that result incurs Undefined Behavior.
I think that if the compiler can prouve that the result is
used (and it is here), then it is allowed to reject the
code. (The case here is an interesting one, in fact---I'm
assuming that the same rules hold for references as for
pointers, and that even copying an invalid reference is
undefined behavior. I've not looked it up, however, so
caveat empor.)
How can compiler prove that the result is used?
By flow analysis.
How can he prove i'm not writing a thread that's supposed to
be externally terminated in the very same case as when the
function happens to return a reference- to-local, the
reference later explicitely being used but unreachable in
practice? :-)
That's undefined behavior as well, as far as the standard goes,
so the compiler doesn't have to take it into account. (In fact,
of course, it's also impossible to do on most modern systems;
another process can't really look at the individual instructions
you're executing.)
--
James Kanze (GABI Software) email:james.kanze@gmail.com
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