Question

In “The C++ Programming Language (3rd)” p.255:

A temporary can be used as an initializer for a const reference or a named object. For example:

void g(const string&, const string&);

void h(string& s1, string& s2)
{
   const string& s = s1+s2;
   string ss = s1+s2;

   g(s, ss);  // we can use s and ss here
}

This is fine. The temporary is destroyed when “its” reference or named object go out of scope.

Is he saying that the temporary object created by s1+s2 is destroyed when ss goes out of scope?
Isn’t it get destroyed as soon as it is copy initialized to ss?

Answer 1

The only temporaries in your code are the s1 + s2. The first one gets bound to the const-ref s, and thus its lifetime is extended to that of s. Nothing else in your code is a temporary. In particular, neither s nor ss are temporaries, since they are manifestly named variables.

The second s1 + s2 is of course also a temporary, but it dies at the end of the line, having been used to initialize ss only.

Update: Perhaps one point deserves emphasis: In the final line, g(s, ss);, the point is that s is a perfectly valid reference, and it is not a dangling reference as you might perhaps have expected, precisely because of the life-time extension rule for temporaries bound to const-references.