当新的运算符被覆盖时，C ++类构造函数被调用两次 - 为什么

Question

我编写了以下代码来解释运算符new

#include <iostream>

using namespace std;




class Dog {

public:
    Dog() {
        cout << "Dog constructed\n";
    }

    //overriding the new operator of the class and explicitely doing what it internally does
    void* operator new(size_t n) {
        cout << "operator new overriden in Dog class called size_t n = " << n << " Sizeof(Dog) = "<< sizeof(Dog) << endl;

        //Allocating the memory  by calling the global operator new 
        void* storage = ::operator new (n);

        // you can now create the Dog object at the allcated memory at the allocated memory

        ::new (storage) Dog(); //---------> Option 1 to construct the Dog object --> This says create the Dog at the address storage by calling the constructor Dog()

        return storage;
    }

    int m_Age = 5;
    int m_Color = 1;
};

void* operator new(std::size_t size)
{
    void* storage = malloc(size);
    std::cout << "Global operator new called - Asked for: " << size
        << ", at: " << storage << std::endl;
    return storage;
}


int main(int argc, char** argv)
{



    cout << "calling new Dog" << endl;
    Dog* ptr = new Dog;
    int x;
    cin >> x;
    return 0;
}

当我运行它时，输出如下

================================================

叫新狗

Dog类中的operator new overriden，名为size_t n = 8 Sizeof（Dog）= 8

全局运营商新召集 - 要求：8，at：0xf15c20

狗构造

狗构造

==========================================

任何想法为什么Dog对象Dog的construcor被调用两次？

谢谢

Answer 1

它是因为你在重载operator new中使用placement new构造一个Dog对象，同时为正在构造的Dog object分配存储：

::new (storage) Dog();

operator new永远不应该构造一个对象，它应该只分配内存，编译器将使用分配的内存发出代码来构造一个对象。

Answer 2

这是因为你要构建Dog在这只能在一定的分配空间功能Dog 。 然后编译器在那个上构造第二个Dog 。

观察返回类型是void* ，即指向unknown的指针。 如果您打算构建一只Dog ，返回值将是Dog*