c++ - 复制和移动构造函数

Question

具有以下代码:

#include <iostream>
#include <vector>
using namespace std;

class CIntPtr {
public:
    int* ptr;
public:
    CIntPtr() {
        // Default constructor
        cout << "Calling Default constructor\n";
        ptr = new int;
    }

    CIntPtr(const CIntPtr& obj) {
        // Copy Constructor
        // copy of object is created
        this->ptr = new int;
        // Deep copying
        cout << "Calling Copy constructor\n";
    }

    CIntPtr(CIntPtr&& obj) {
        // Move constructor
        // It will simply shift the resources,
        // without creating a copy.
        cout << "Calling Move constructor\n";
        this->ptr = obj.ptr;
        obj.ptr = NULL;
    }

    ~CIntPtr() {
        // Destructor
        cout << "Calling Destructor\n";
        delete ptr;
    }

};

CIntPtr returnCIntPtr(CIntPtr a) {
    *a.ptr += 2;
    return a;
}

int main() {

    CIntPtr foo;
    returnCIntPtr(foo);
    
    return 0;
}

同时存在 复制和 移动构造函数。为什么还要调用move ctor？因为我们将左值作为函数参数传递了，难道不是仅调用了复制ctor吗？何时可以在此代码中调用move ctor？

Answer 1

在return a;函数的return statement(即returnCIntPtr)中调用了移动构造器。

Automatic move from local variables and parameters

If expression is a (possibly parenthesized) id-expression that names avariable whose type is either

• a non-volatile object type or
• a non-volatile rvalue reference to object type (since C++20)

and that variable is declared

• in the body or
• as a parameter of

the innermost enclosing function or lambda expression,

then overload resolution to select the constructor to use for initialization of the returned value or, for co_return, to select the overload of promise.return_value() (since C++20) is performed twice:

• first as if expression were an rvalue expression (thus it may select the move constructor), and
  • if the first overload resolution failed or
• then overload resolution is performed as usual, with expression considered as an lvalue (so it may select the copy constructor).

首先尝试将 a复制为返回值作为rvalue-expression，然后选择move构造函数。