c++ - 在 C++ 中显式调用构造函数和析构函数

Question

我知道不应该为本地成员显式调用析构函数，因为它会导致未定义的行为。但我仍然想了解以下代码的输出:

#include <iostream>
using namespace std;

class Test
{
public:
    Test() { cout << "Constructor is executed\n"; }
    ~Test() { cout << "Destructor is executed\n"; }
    friend void fun(Test t);
};
void fun(Test t)
{
    Test(); //3.calls constructor and destructor
    t.~Test(); //4. calls destructor
}
int main()
{
    Test(); // 1. calls constructor and destructor
    Test t; // 2.calls constructor
    fun(t); 
    return 0; //5. compiler calls destructor for t
}

预期输出(最后调用 3 个析构函数):

Constructor is executed
Destructor is executed
Constructor is executed
Constructor is executed
Destructor is executed
Destructor is executed
Destructor is executed

实际输出(最后调用了 4 次析构函数):

Constructor is executed
Destructor is executed
Constructor is executed
Constructor is executed
Destructor is executed
Destructor is executed
Destructor is executed
Destructor is executed

最后额外的析构函数调用在哪里？

Answer 1

您按值将 t 传递给 fun。它创建一个拷贝。你看不到的原因是你没有让复制构造函数打印任何东西。但你确实看到它被破坏了。