Neither Default Constructor nor Copy Constructor is called when using anonymous object

Question

Below is the snippet aimed to test the constructors. It was run in VS 2015.

In my opinion, "B b(B())" has the same function as "B b = B()", however, my code seems to say that they behave different.

I know there's copy elision by compiler optimization, but I think the default constructor should be called at least when execute "B b(B())". Can anyone help point out where my misunderstood is?

class B
{
public:
    B() {
        ++i;
        x = new int[100];
        cout << "default constructor!"<<" "<<i << endl;
        cout << "x address:" << x << endl << "--------" << endl;
    }
    B(const B &b)  //copy constructor
    {   
        ++i;
        cout << "Copy constructor & called " << i<< endl 
            << "--------" << endl;
    }
    B(B &&b)//move constructor
    {
        x = b.x;    
        ++i;
        b.x = nullptr;
        cout << "Copy constructor && called" << i << endl 
            <<"x address:"<< x << endl << "--------" << endl;

    }
    void print()
    {
        cout << "b address:" << x << endl << "--------" << endl;
    }
private:
    static int i;
    int *x;
};
int B::i = 0;


int main()
{
    B b1; //default constructor
    b1.print();
    B b2 = B(); //default constructor only
    b2.print();
    B b3(B());  //????nothing called... why???
    B b4 = b2; //copy constructor
    B b5 = move(b2); //move constructor
    b2.print();

    return 0;
}

songyuanyao · Accepted Answer · 2016-10-18T03:11:03.200

4

Note that B b(B()) is a function declaration, not a variable definition at all, then no constructor would be called.

According to Most vexing parse, B b(B()) is a function declaration, for a function named b which returns an object of type B and has a single (unnamed) parameter which is a pointer to function returning type B and taking no parameter.

You can solve it by using braces (list initlization (since C++11)), like

B b1( B{} );
B b2{ B() };
B b3{ B{} };  // preferable from C++11

edited Oct 18 '16 at 03:11

answered Oct 18 '16 at 02:49

songyuanyao

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Nice catch! @Shirley Feng, according to the wiki article, you can get the expected behavior by adding parenthesis around your inner ctor call. Example: `B b3( (B()) );` – souldzin Oct 18 '16 at 02:55
Thank you for the share!! It saves me XD – Shirley Feng Oct 18 '16 at 02:57
Just wonder, shouldn't it be better to keep using one convention for initialization? i.e. `B b{ B{} };` instead – Adrian Shum Oct 18 '16 at 03:07
1

@AdrianShum You're right. I just want to point out that only using one pair of braces would solve the ambiguity here. – songyuanyao Oct 18 '16 at 03:10
@ShirleyFeng Yes! It's a really important C++11 feature. – songyuanyao Oct 18 '16 at 03:31
@songyuanyao Hi, I have a new doubt about the call of copy constructor regarding anonymous object. B b1{ move(B{}) }; can call the move constructor. but is there anyway that we can trigger the copy constructor when initialize an object with another anonymous one? Thanks! – Shirley Feng Oct 18 '16 at 05:39
@ShirleyFeng What do you want to achieve? You can write a function similar with `std::move`, like `B foo(const B& b) { return b; }`, then `B b{ foo(B{}) };`. – songyuanyao Oct 18 '16 at 06:34

Neither Default Constructor nor Copy Constructor is called when using anonymous object

1 Answers1