C++ object creation and constructor

Question

I'm learning ctors now and have a few question. At these lines:

Foo obj(args);

Foo obj2;
obj2 = Foo(args);

Foo obj3 = Foo(args);

First part: only 1 constructor called (Foo) and obj is initialized. So, 1 object creation.

Second part: creating of temporary object obj2, calling default ctor for it. Next lines we create another copy of Foo and pass its copy into operator=(). Is that right? So, 3 local temporary objects, 2 constructor callings.

Third part: create 1 object Foo and pass its copy into operator=(). So, 2 temprorary objects and 1 ctor calling.

Do I understand this right? And if it's true, will compiler (last gcc, for example) optimize these in common cases?

`operator=()` usually receives its argument by reference, so no copies in calling it. — rodrigo, Jan 08 '12 at 11:59
obj3 is actually created using copy constructor not assignement operator. — Ivaylo Strandjev, Jan 08 '12 at 11:59

Nawaz · Accepted Answer · 2012-01-08T12:04:47.523

16

I will comment on the third one first:

Foo obj3=Foo(args);

It doesn't use operator= which is called copy-assignment. Instead it invokes copy-constructor (theoretically). There is no assignment here. So theoretically, there is two objects creation, one is temporary and other is obj3. The compiler might optimize the code, eliding the temporary object creation completely.

Now, the second one:

Foo obj2;         //one object creation
obj = Foo(args);  //a temporary object creation on the RHS

Here the first line creates an object, calling the default constructor. Then it calls operator= passing the temporary object created out of the expression Foo(args). So there is two objects only the operator= takes the argument by const reference (which is what it should do).

And regarding the first one, you're right.

edited Jan 08 '12 at 12:04

answered Jan 08 '12 at 11:59

Nawaz

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Okay, thanks. But anyway, in first case only 1 objects of type `Foo` is created, in third one: 2 objects? – Max Frai Jan 08 '12 at 12:01
No, according to the specs, part 1 and part 3 are two ways of specifying the same thing. There is no difference in implementation. – Mr Lister Jan 08 '12 at 12:11
4

@MrLister: No. There is a subtle difference between `1` and `3`. Just write a test code, and make the copy-constructor `private`. The 1st one will compile, the third one will not! – Nawaz Jan 08 '12 at 12:21
Aw. I'm sure I was taught this when I learned C++. But you're right that it won't compile. And I can't find references to it being synonyms for the same thing. So, thanks and sorry. And I'm gonna ask my money back. – Mr Lister Jan 08 '12 at 12:51

score 3 · Answer 2 · answered Jan 08 '12 at 12:01

Yes, Foo obj(args) creates one Foo object and calls the ctor once.
obj2 is not considered a temporary object. But just like 1 Foo obj2 creates one object and calls the Foo ctor. Assuming that you meant obj2 = Foo(args) for the next line, this line creates one temporary Foo object and then calls obj2.operator=(). So for this second example there is only a single temporary object, a single non-temporary, Foo ctors are called twice (once for the non-temporary, once for the temporary) and the operator=() is called once.
No, this line does not call operator=(). When you initialize obj3 using the = syntax it is almost exactly as if you had used parentheses instead: Foo obj3(Foo(args)); So this line creates a temporary object, and then calls the Foo copy ctor to initialize obj3 using that temporary object.

score 2 · Answer 3 · answered Jan 08 '12 at 12:08

Your terminology is a bit confusing.

The objects obj, obj2 obj3 are not called "temporary objects". Only the instance that is created in line 3 before being assign to obj is a temporary object.

Also, you don't create "a copy of Foo", you create either "an instance of Foo" or "an object of type Foo".

C++ object creation and constructor

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