Thread: Help me understand copying better

It doesn't. It uses a built-in assignment operator instead.

No, it uses the compiler provided copy constructor.

In that case the compiler is allowed to also use the copy-constructor, instead of a call to the default constructor followed by a call to the assignment operator. So you can't actually say that it will call the assignment operator. It's compiler dependent.

I do not have a copy of the C++ Standard with me at the moment, but I believe that it is not compiler dependent: this syntax means an implicit invocation of the copy constructor. If you declare the copy constructor explicit, then one must use:

Code:

SomeObject object2(object1);

instead of:

Code:

SomeObject object2 = object1;

Originally Posted by laserlight

I do not have a copy of the C++ Standard with me at the moment, but I believe that it is not compiler dependent: this syntax means an implicit invocation of the copy constructor.

You could be right. I can't confirm it either.