Question

template <class Type> class Queue {


    Queue(): head(0), tail(0) {
        cout << "Queue--default constructor called" << endl;
    }

    Queue(const Queue &Q): head(0), tail(0) {
        cout << "Queue--copy constructor called" << endl;
        //...
    }

    Queue& operator=(const Queue&) {
        cout << "Queue--operator= called" << endl;
        //...
    }
    ~Queue() { //... }

private:
    QueueItem<Type> *head;         
    QueueItem<Type> *tail;         
};

I have defined a template class Queue and tried the codes below:

Queue<char*> cq;
Queue<char*> ccq(cq);   
Queue<char*> acq = cq;  
Queue<char*> acq2;
acq2 = cq;

and the output is:

Queue--default constructor called
Queue--copy constructor called
Queue--copy constructor called
Queue--default constructor called
Queue--operator= called

what confuse me is the code Queue<char*> acq = cq; has invoked the copy constructor Queue--copy constructor called but not the default constructor and operator= to be called.

Could any one help me?

Thank you for considering my question!

Answer 1

This is defined in the standard, it’s in the section Explicit initialization


1
Explicit initialization
[class.expl.init]
An object of class type can be initialized with a parenthesized expression-list, where the expression-list
is construed as an argument list for a constructor that is called to initialize the object. Alternatively, a
single assignment-expression can be specified as an initializer using the = form of initialization. Either
direct-initialization semantics or copy-initialization semantics apply; see 8.5.

This is the case describing single assignment-expression and copy-initialization semantics.

Technically there’s no difference between the two forms ccq(cq) and acq = cq, as you have already seen from the output.