c++ - 从函数和复制构造函数返回的对象

Question

有这样的代码:

#include <iostream>

class A {
public:
    int a;
    A() : a(0) {
        std::cout << "Default constructor" << " " << this << std::endl;
    }
    A(int a_) : a(a_) {
        std::cout << "Constructor with param " << a_ << " " << this << std::endl;
    }
    A(const A& b) {
        a = b.a;
        std::cout << "Copy constructor " << b.a << " to " << a << " " << &b << " -> " << this << std::endl;
    }
    A& operator=(const A& b) {
        a=b.a;
        std::cout << "Assignment operator " << b.a << " to " << a << " " << &b << " -> " << this <<  std::endl;
    }
    ~A() {
        std::cout << "Destructor for " << a << " " << this << std::endl;
    }
    void show(){
      std::cout << "This is: " << this << std::endl;
    }
};


A fun(){
  A temp(3);
  temp.show();
  return temp;
}


int main() {
    {
      A ob = fun();
      ob.show();
    }
    return 0;
}

结果:

Constructor with param 3 0xbfee79dc
This is: 0xbfee79dc
This is: 0xbfee79dc
Destructor for 3 0xbfee79dc

对象 ob 由函数 fun() 初始化。为什么不在那里调用复制构造函数？我认为当函数按值返回时，将调用复制构造函数或赋值运算符。似乎在函数 fun() 中构造的对象在函数执行后没有被销毁。在这种情况下，如何强制调用复制构造函数？

这是由 g++ 编译的。

Answer 1

Why copy constructor is not called there?

RVO

How can be copy constructor forced to invoke in this case?

传递一个选项给编译器。对于 gcc，它是 --no-elide-constructors 选项来禁用 RVO