Guaranteed lifetime of temporary in C++?

Question

Does C++ provide a guarantee for the lifetime of a temporary variable that is created within a function call but not used as a parameter? Here\'s an example class:

Answer 1

litb's answer is accurate. The lifetime of the temporary object (also known as an rvalue) is tied to the expression and the destructor for the temporary object is called at the end of the full expression and when the destructor on StringBuffer is called, the destructor on m_buffer will also be called, but not the destructor on m_str since it is a reference.

Note that C++0x changes things just a little bit because it adds rvalue references and move semantics. Essentially by using an rvalue reference parameter (notated with &&) I can 'move' the rvalue into the function (instead of copying it) and the lifetime of the rvalue can be bound to the object it moves into, not the expression. There is a really good blog post from the MSVC team on that walks through this in great detail and I encourage folks to read it.

The pedagogical example for moving rvalue's is temporary strings and I'll show assignment in a constructor. If I have a class MyType that contains a string member variable, it can be initialized with an rvalue in the constructor like so:

class MyType{
   const std::string m_name;
public:
   MyType(const std::string&& name):m_name(name){};
}

This is nice because when I declare an instance of this class with a temporary object:

void foo(){
    MyType instance("hello");
}

what happens is that we avoid copying and destroying the temporary object and "hello" is placed directly inside the owning class instance's member variable. If the object is heavier weight than a 'string' then the extra copy and destructor call can be significant.