Copy constructor and assignment operator invocation cases

In many cases, it is difficult to accept the following instructions

int a (10);

is actually exactly equivalent to int a=10;

So the following situation is more difficult to accept:

Class a;//defines class A

Application:

A;

A B (a);//is actually equivalent to a b=a, the call is copy constructor, in everyone's eyes if not, very familiar with the way of such assignment, it is really considered an assignment operation.

Examples are as follows:

void Test (A A)

{

}

In this function, by passing the value, the copy constructor is actually called, not the assignment operator, or the default constructor.

Class t{public:    t ()     {        cout<< "T ()" <<endl;    }    t (const t&  T)     {        cout<< "call copy  Constructor "<<endl;    }    t operator = (T t)     {        cout<< "call assign  Constructor "<<endl;    }    ~t ()     {        cout<< "~t ()" <<endl;    }private:};void The detailed invocation of   test (t t) {} is as follows:         t *p=new t ();         t c=*p;        t  cc;        cc=*p;        cout<< "* *" <<endl ; &NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;&NBSP;T&NBSP;BB (CC);         cout<< "* * *" <<endl;        test (*p);         delete p;

Copy constructor and assignment operator invocation cases