temporary object, function parameters and implicit cast

Question

In the following scenario:

struct Foo
{
   // ...
   operator Bar() {... }  // implicit cast to Bar
}

Foo GetFoo() { ... }
void CallMeBar(Bar x) { ... }

// ...         


        

Answer 1

Regardless of the cast, the temporary object(s) will "survive" the call to CallMe() function because of the C++ standard:

12.2.3 [...] Temporary objects are destroyed as the last step in evaluating the fullexpression (1.9) that (lexically) contains the point where they were created. [...]

1.9.12 A fullexpression is an expression that is not a subexpression of another expression.