lifetime extension for function taking parameter by const& and returning by const&

Question

In C++ when you have the following:

std::string get_string();
std::string const& v = get_string();

The lifetime of the temporary returned from get_string() is extended for the same lifetime of the reference v;

If I have the following:

std::string const& get_string(std::string const& p) {
  return p;
}

std::string const& v = 
get_string(std::string{"Hello"});

Is the lifetime of the temporary extended? or is this a dangling reference;

My understanding is that the temporary is bound to the lifetime of p and that only exists for the duration of the function and that secondary references to the temporary dont extend the lifetime.

What is the expected result?

Answer 1

Yes, the lifetime of the temporary is not extended further more; after the full expression the reference v becomes dangled.

std::string const& v = get_string(std::string{"Hello"});
// v becomes dangled now

My understanding is that the temporary is bound to the lifetime of p and that only exists for the duration of the function

To be precisely, the temporary exists until the end of the full expression, not only the duration of the function.

• a temporary bound to a reference parameter in a function call exists until the end of the full expression containing that function call: if the function returns a reference, which outlives the full expression, it becomes a dangling reference.

In general, the lifetime of a temporary cannot be further extended by "passing it on": a second reference, initialized from the reference to which the temporary was bound, does not affect its lifetime.

That means something like auto sz = get_string(std::string{"Hello"}).size(); is fine.