C++ copy constructor invocation

Question

As far as i know, a copy constructor is invoked in the following scenarios :

1) Pass by value
2) Return by value
3) When you create and initialize a new obje         


        

Answer 1

In your exampe the structure is small enough therefore it is passed through a register. The generated code is similar to Return value optimization. Construct a more complicated example, and you'll see the behavior expected.

Answer 2

The part of the standard which tells you when compilers may elide copies is 12.8/15. It's always up to the compiler whether to do actually perform the elision. There are two legal situations, plus any combination of them:

• "in a return statement in a function with a class return type, when the expression is the name of a non-volatile automatic object with the same cv-unqualified type as the function return type"

• "when a temporary class object that has not been bound to a reference would be copied to a class object with the same cv-unqualified type".

The former is usually referred to as the "named return value optimization", and it's what permits the output you're seeing in your example. The latter in effect turns copy-initialization into direct initialization, and could occur for instance if your code did Example x = Example();.

Other copy elisions are not permitted, except of course that the usual "as-if" rules apply. So if the copy constructor has tracing in, then the following code must call it:

Example x;
Example y = x;

But if x were otherwise unused, and the cctor had no side-effects, then I think it could be optimized away, just like any other code that does nothing.

Answer 3

When doing x = funct(); the compiler notices that it will be directly returned and thus avoids a useless construction. That's also why you will only get two destructor calls.

This is a example why sometimes working with "copy" isn't necessarily a lost of performances.

Answer 4

A copy constructor might not be invoked when you return by value. Some compilers use return value optimization feature.

Read about "Return Value Optimization"

Answer 5

g++ v4.4.1 has an option to suppress "elide" optimizations:

tst@u32-karmic$  g++ -fno-elide-constructors Example.cpp -o Example

tst@u32-karmic$  ./Example 

Default constructor called.
Calling funct..
Default constructor called.

Copy constructor called.

Dtor invoked
Assignment operator called.

Dtor invoked

Dtor invoked

As you can see the copy constructor is now called!