Parameter passed by const reference returned by const reference

Question

I was reading C++ Faq Second Edition , faq number 32.08 .

FAQ says that parameter passed by const reference and returned by const reference can cause dangling reference.

But it is ok if parameter is passed by reference and returned by reference.

I got it that it is unsafe in case of const reference but how is it safe in case when parameter is non const reference.

Last line of FAQ says "Note that if a function accepts a parameter by non-const reference (for example, f(string& s)), returning a copy of this reference parameter is safe because a temporary cannot be passed by non-const reference."

Need some insight on this!!

Answer 1

if you have like

const Foo & bar(const Foo &f) { return f; }

and call it like

const Foo &ret = bar(Foo());

This compiles, but the problem is that now 'ret' is a dangling reference, because the temporary object created by the call to Foo() gets freed after bar returns. The detailed execution sequence here is:

1. temporary Foo is allocated
2. bar is called with a reference to the temporary object
3. bar returns the reference
4. now that bar has returned the temporary Foo is released
5. the reference is now dangling as the object was destroyed

However, if you had Foo declared as

Foo & bar(Foo &f) { return f; }

then your call bar(Foo()) would not be accepted by compiler. When you pass a temporary object to a function, you can only take it by const reference or as a copy; this is part of the language definition.

Answer 2

Temporaries can be passed by const reference - when the function returns the temporaries are destoyed, so the caller is left with a dangling ref.

For example:

#include <iostream>

using namespace std;

int const& output( int const& x) 
{
    cout << x << endl;

    return x;
} 

int main ()
{
    int a = 1;

    int const& ref1 = output( a); // OK

    int const& ref2 = output(a+1); // bad

    return 0;
}