Why there is no temporary object when returning an object through the constructor?

Question

I'm trying to figure out what exactly happens when returning an object through the constructor(conversion function).

Stonewt::Stonewt(const Stonewt & obj1) {
    cout << "Copy constructor shows up." << endl;
}

Stonewt::Stonewt(double lbs) {
    cout << "Construct object in lbs." << endl;
}

Stonewt::~Stonewt() {
    cout << "Deconstruct object." << endl;
}

Stonewt operator-(const Stonewt & obj1, const Stonewt & obj2) {
    double pounds_tmp = obj1.pounds - obj2.pounds;
    return Stonewt(pounds_tmp);
}

int main() {
    Stonewt incognito = 275;
    Stonewt wolfe(285.7);

    incognito = wolfe - incognito;
    cout << incognito << endl;
    return 0;
}

Output:
Construct object in lbs.
Construct object in lbs.

Construct object in lbs.
Deconstruct object.
10.7 pounds

Deconstruct object.
Deconstruct object.

So My Question is:

Why there is no copy constructor (no temporary object) when returning an object through the constructor?

What exactly are you asking? Do you just not understand what a destructor is? You have too much code here to expect a decent answer as-is. — Ryan Haining, Mar 28 '18 at 03:03
I think you're looking for [copy elision](http://en.cppreference.com/w/cpp/language/copy_elision) — Ryan Haining, Mar 31 '18 at 02:22
Possible duplicate of [What are copy elision and return value optimization?](https://stackoverflow.com/questions/12953127/what-are-copy-elision-and-return-value-optimization) — Ryan Haining, Mar 31 '18 at 02:35

Sid S · Accepted Answer · 2018-03-29T17:50:33.513

3

Stonewt operator-(const Stonewt & obj1, const Stonewt & obj2)
{
    ...
    return obj1;
}

   incognito = incognito - wolfe;

Your operator - () is returning a copy of incognito, which you then assign to incognito. The copy is then destroyed.

edited Mar 29 '18 at 17:50

answered Mar 28 '18 at 03:09

Sid S

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I thought the temporary object was created by the defaul constructor, so I added "cout << "Construct object with defaul constructor." << endl;" in the defalt constructor function. But there still no information about the constructor in the output. So which constructor it uses to create the temporary object? Thank you~ – Guodong Hu Mar 28 '18 at 06:37
@LeoHu, that's because the default copy constructor is used. Implement a copy constructor, and it will be called. – Sid S Mar 28 '18 at 20:25

Why there is no temporary object when returning an object through the constructor?

1 Answers1