c++ - 为什么销毁顺序与构造顺序相同，静态对象(C++)？

Question

代码是:

#include <iostream>
using namespace std;

class A
{
public:
    A() { cout << "A::A" << endl; }
    ~A() { cout << "A::~" << endl; }
};

class B
{
public:
    B() { cout << "B::B" << endl; }
    ~B() { cout << "B::~" << endl; }
};

int main()
{
    B b;
    static A a;
    return 0;
}

输出是:

B::B
A::A
B::~
A::~

非静态对象 b 的范围和静态对象 a 的范围在 main() 函数结束时结束。

问题:为什么构造函数的顺序和析构函数的顺序一样？

Answer 1

Static local variables将在程序退出时销毁。

The destructor for a block-scope static variable is called at program exit, but only if the initialization took place successfully.

所以b会在main()结束时先被销毁，a会在这之后被销毁。

对于初始化，

are initialized the first time control passes through their declaration

所以b会在main()中先被初始化，然后a被初始化。