自动销毁对象，即使在显式调用析构函数之后也是如此

问题描述

以下程序:

#include <iostream>
using namespace std;

class Test
{
public:
    Test() { cout << "Constructor is executed\n"; }
    ~Test() { cout << "Destructor is executed\n"; }
};

int main()
{
    Test(); // Explicit call to constructor
    Test t; // local object
    t.~Test(); // Explicit call to destructor
    return 0;
}

打印以下输出:

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

我的问题甚至是在main()中显式调用析构函数之后，为什么编译器在退出main()之前隐式调用析构函数?

My question is even after explicitly calling destructor in main(), why does the compiler call the destructor implicitly before exiting main()?

作为附带的问题，除了在delete运算符中使用之外，显式调用析构函数的策略还有其他用途吗?

As a side question, apart from use in delete operator is there any other use of the strategy of calling destructor explicitly?

推荐答案

您已经引入了未定义的行为.

You've introduced undefined behavior.

按照标准:

(11)析构函数被隐式调用

(11) A destructor is invoked implicitly

(11.3)— 用于当对象所在的块所在时具有自动存储持续时间(3.7.3)的构造对象 创建的出口(6.7)，

(11.3) — for a constructed object with automatic storage duration (3.7.3) when the block in which an object is created exits (6.7),

和

您显式调用析构函数或通过调用t.~Test()，然后在对象离开作用域时隐式调用它.这是未定义的.

You explicitly call the destructor or by calling t.~Test(), it is then implicitly invoked when the object leaves scope. This is undefined.

该标准也提供了此注释:

The standard provides this note as well:

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