Question

If a local object is returned in a function call, it has to do at least three steps:

1. Copy constructor is called to hold a copy.
2. Destroy local object.
3. A copy is return.

For example:

x = y + z

If x is an integer object. A copy of y + z should be returned, then a new object is created, then assignment operator of x will take this object as parameter.

So my questions are:

• Is the same process used for built-in type such as int, double…?
• If they’re not the same, how’s it done?

Answer 1

The language specification does not say “how it is done” for built-in types and built-in operators. The language simply says that the result of binary + for built-in types is an rvalue – the sum of the operand values. That’s it. There’s no step-by-step description of what happens when a built-in operator is used (with some exceptions like &&, , etc.).

The reason you can come up with a step-by-step description of how an overloaded operator works (which is what you have in your question) is because the process of evaluating overloaded operator comes through several sequence points. A sequence point in a C++ program implements the concept of discrete time: it is the only thing that separates something that happens before from things that happen after. Without a separating sequence point, there’s no “before” and no “after”.

In case of an overloaded operator, there quite a few sequence points involved in the process of its evaluation, which is why you can describe this process as a sequence of steps. The process of evaluation of built-in operator + has no sequence points in it, so there absolutely no way to describe what happens there in step-by-step fashion. From the language point of view, the built-in + is evaluated through a blurry indivisible mix of unspecified actions that produce the correct result.

It is done this way to give the compiler better optimization opportunities when evaluating built-in operators.