Return reference from function vs binding temporary to const ref

Question

Am I right, that this (taken from this GotW):

string f() { return "abc"; }

void g() {
    const string& s = f();
    cout << s << endl;    // can we still use the "temporary" object?
}

is completely fine, while this is not ok at all:

const string& foo() { string x{"abc"}; return x; }
void bar() {
    const string& y = foo();
}

?

Why is it different? Can one say that "the lifetime of a temporay is not extended across a function call (even if bound to a const ref)" ? Or what is an explanation, why the first is ok, but not the second?

First answer there answers your questions. – Hatted Rooster May 19 '17 at 09:49 — Hatted Rooster, May 19 '17 at 09:49
@GillBates yep I also just found it – 463035818_is_not_an_ai May 19 '17 at 09:49 — 463035818_is_not_an_ai, May 19 '17 at 09:49

songyuanyao · Answer 1 · 2017-05-19T09:56:25.930

The 1st one is pretty fine, because the returned value (which is a temporary) is bound to a reference to const directly, then the lifetime of the temporary is extended to the reference's lifetime.

Note that there's no temporary in the 2nd case, x is a local variable which will be destroyed when get out of the function, then the function will always returns a dangled reference; that reference is used to initialize y but nothing changes, y will be a dangled reference too.

Return reference from function vs binding temporary to const ref

1 Answers1