Order of constructor and destructor calling?

问题

I cannot understand the order of constructor and destructor calls? What will execute first in this statement A b=f(a)? Can someone please help me out?

#include<iostream>
using namespace std;

class A {
    int x;

    public:
        A(int val = 0)
        :x(val) {
            cout << "A " << x << endl << flush;
        }
        A(const A& a) {
            x = a.x;
            cout << "B " << x << endl << flush;
        }
        void SetX(int x) {
            this->x = x;
        }
        ~A() {
            cout << "D " << x << endl << flush;
        }
};

A f(A a) {
    cout << " C " << endl << flush;
    a.SetX(100);
    return a;
}

int main()
{
    A a(1);
    A b=f(a);
    b.SetX(-100);
    return 0;
}

Output Window:

A 1
B 1
 C
B 100
D 100
D -100
D 1

Why does it print B 1 in line 2 of the output window?

回答1:

"Why does it print B 1 in line 2?"

Because the copy constructor was called from this statement

A b=f(a);

The function f() requires A being passed by value, thus a copy for this parameter is made on the function call stack.

---

If your next question should be, how you can get over this behavior, and avoid to call the copy constructor, you can simply pass the A instance as a reference to f():

    A& f(A& a) {
  // ^    ^
        cout << " C " << endl << flush;
        a.SetX(100);
        return a;
    }

---

_{Side note: endl << flush; is redundant BTW, std::endl includes flushing already.}

来源：https://stackoverflow.com/questions/29854010/order-of-constructor-and-destructor-calling