why can't print twice deconstruct for this C++ code sample?

Question

using namespace std;

Object returnObject(){
    Object o;  
    return o;  //place A
 }

int main() {
    Object CopiedO=returnObject();  
    return 0;  //Place B
}

the object definition is:

Object::Object() {
    cout<<"Object::Object"<<endl;
}

Object::~Object() {
    cout<<"Object::~Object"<<endl;
}

Object::Object(const Object& object) {
    cout<<"Object::CopyObject"<<endl;
}

the result is:

/*Object::Object
Object::~Object*/

As I understand, both o and CopiedO in will be deconstructed, but why only once Object::~Object be printed?

I think there is no inline and the copied o is copy of o .but it can't print Object::CopyObject

This is a fine example of return value optimisation (RVO) at work. :-) — Chris Jester-Young, Jan 10 '14 at 04:32
Removed my answer saying it was move semantics. That would not be the case because the default move constructor is not automatically generated when you have a user-defined copy constructor. — Mark, Jan 10 '14 at 04:51
You can try split the assignment into two lines: `Object CopiedO; CopiedO = returnObject();` and you will see the difference how the optimization works. — Mine, Jan 10 '14 at 05:25

score 4 · Accepted Answer · answered Jan 10 '14 at 04:35

4

The compiler is eliding the copy. Since it knows that the returned object from the function has as only purpose to initialize CopiedO, it is merging both objects into one and you will see only one construction, one destruction and no copy-constructions.

answered Jan 10 '14 at 04:35

David Rodríguez - dribeas

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why can't print twice deconstruct for this C++ code sample?

1 Answers1