在C ++中显式调用构造函数和析构函数

Question

I know that one should not call destructors explicitly for local members as it results in undefined behaviour. But still I want to understand the output for following piece of code :

#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
}

Expected output (3 destructor calls at the end):

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

Actual output (4 destructor calls at the end):

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

Where is th extra destructor call coming at the end?

Answer 1

You pass t to fun by value. It creates a copy. The reason you don't see that is because you didn't make the copy constructor print anything. But you do see it destructed.

Answer 2

变量t通过值传递给函数fun - 因此在函数入口处构造了一个额外的副本（您没有看到这个打印，因为您的print语句仅用于默认构造函数），并且在退出时也被破坏（这是在t上，您还调用析构函数明确，所以它被破坏的两倍）。

Answer 3

You should at least print address of this in ctor and dtor message. I did it and got:

Constructor is executed 0020FC9F
Destructor is executed0020FC9F
Constructor is executed 0020FD77
Constructor is executed 0020FBAB
Destructor is executed0020FBAB
Destructor is executed0020FC84
Destructor is executed0020FC84
Destructor is executed0020FD77

This shows that you get an extra object at 20FC84 on which you call destructor twice without seeing it in default constructor. It is because it has been created with an implicit (compiler provided) copy constructor to allow it to be passed by value to function fun . If you want to avoid that extra copy just declare fun to take its argument by ref:

void fun(Test& t)
{
    Test(); //3.calls constructor and destructor
    t.~Test(); //4. calls destructor
}

(do not forget the ref in the friend declaration...)

It now gives as expected:

Constructor is executed 003EF90F
Destructor is executed003EF90F
Constructor is executed 003EF9E7
Constructor is executed 003EF827
Destructor is executed003EF827
Destructor is executed003EF9E7
Destructor is executed003EF9E7

Answer 4

In this statement in main

Test(); 

the constructor and the destructor of the temporary object is called.

So you have

Constructor is executed Test() in main
Destructor is executed  Test() in main
Constructor is executed
Constructor is executed
Destructor is executed
Destructor is executed
Destructor is executed
Destructor is executed

Then in the next statement in main

Test t;

the constructor is called and you have

Constructor is executed Test() in main
Destructor is executed  Test() in main
Constructor is executed Test t in main
Constructor is executed
Destructor is executed
Destructor is executed
Destructor is executed
Destructor is executed

Then in this statement in fun

Test(); 

the constructor and the destructor of the temporary object are called and you have

Constructor is executed Test() in main
Destructor is executed  Test() in main
Constructor is executed Test t in main
Constructor is executed Test() in fun
Destructor is executed  Test() in fun
Destructor is executed
Destructor is executed
Destructor is executed

Then the destructor is called explicitly in fun

t.~Test(); 

and you have

Constructor is executed Test() in main
Destructor is executed  Test() in main
Constructor is executed Test t in main
Constructor is executed Test() in fun
Destructor is executed  Test() in fun
Destructor is executed  t.~Test() in fun
Destructor is executed
Destructor is executed

In turn the compiler implicitly calls the destructor for its parameter t after exiting the function

Constructor is executed Test() in main
Destructor is executed  Test() in main
Constructor is executed Test t in main
Constructor is executed Test() in fun
Destructor is executed  Test() in fun
Destructor is executed  t.~Test() in fun
Destructor is executed  The compiler calls the destructor for parameter t
Destructor is executed

And at last the local object t declared in main is also destroyed after exiting main

Constructor is executed Test() in main
Destructor is executed  Test() in main
Constructor is executed Test t in main
Constructor is executed Test() in fun
Destructor is executed  Test() in fun
Destructor is executed  t.~Test() in fun
Destructor is executed  The compiler calls the destructor for parameter t
Destructor is executed  The compiler calls the destructor for local object t declared in main

Answer 5

最后一个析构函数用于fun（），它占用了堆栈空间析构函数在所有当前线程完成执行后执行，它删除了函数分配的空间。