Question

Multi thread enviroment . The content of Foo can be multi thread.

class Foo
{
public:
   const A & getA() {return a_;} //has guard
   void setA(A newA){a_ = newA;} //has guard

private:
    A a_;
};

caller:

A a  = foo.getA();

in another question that i asked someone told me that
If you return const& it’s guaranteed that life time of variable will be prolonged to lifetime of the reference
, so according to this i dont need to copy the value and i safe even if call to setA on foo done right after i call to getA, but a lot of argument against it was araised , so i feel that this is not correct.

I want to be on the safe side so i change the signature to be :

A & getA() {return a_;}

but the compiler still give me warning that i have reference to local variable. i dont understand why, because as far as i understand (new to cpp) the return value is a copy of foo.a, so what the problem with this?

i am not worried about change of a_ content.(_a.age =4) . i worry about call to set and that my ‘a’ in the caller will be illegal

Answer 1

You need to be more careful who you listen to. The only time the lifetime of something gets extended if a temporary object is bound immediately to a const-reference. For example, like so:

Foo bar() { return Foo(); }

int main()
{
    Foo const & f = bar();

    /* stuff */

} // the object referred to by f is extended till here

Your situation is nothing like that. In particular, returning a const-reference does not create a temporary object, so there’s nothing here who’s life gets prolonged. In particular, the following is definitely an error:

A const & bar() { Foo x; return x.getA(); }

int main()
{
    A const & a = bar(); // dangling reference; object dies upon return from Foo::getA()
}