Re: copy constructor and assignment

From:

Bart van Ingen Schenau <bart@ingen.ddns.info>

Newsgroups:

comp.lang.c++.moderated

Date:

Tue, 15 Dec 2009 10:10:51 CST

Message-ID:

<90e87f33-fc67-4c0f-bd49-5cf0577d8fb3@g26g2000yqe.googlegroups.com>

On Dec 15, 1:31 am, samhas <sahaselho...@googlemail.com> wrote:

Hello,

I have the following class definition:

class test2{
public:
  int y;
  test2(int ui) : y(ui){}
  test2(const test2& t) : y(5){}
  test2& operator= (const test2& t){ y = 6; return *this; }

};

In an expression like

test2 t2(test2(20)); or
test2(t2) = test2(20);

neither the copy-constructor nor the assignment operator gets
executed.
t2 ends up with y .

But in the following case the copy constructor gets called:

test2 t = test2(20);
test2 t2(t); // or test2 t2 = t;

Can someone explain to me what's going on there?

First, even though some declarations use the =-sign, it is not an
assignment, so the assignment operator will not be used.

In the declaration 'test2 t2(test2(20));', you are constructing a new
object from a temporary object. For this situation, the compiler has a
special dispensation that it can eliminate the temporary object and
directly initialise the named object from the source of the temporary
object even if that optimisation causes an observable effect.
So, your compiler is interpreting it as being identical to 'test2 t2
(20);' and is fully withing its rights to do so.

Bart v Ingen Schenau

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