c++ - 析构函数必须与构造函数对静态成员的操作相反？

Question

class A
{
    int id;
    static int count;
public:
    A()
    {
        count++;
        id = count;
        cout << "constructor called " << id << endl;
    }
    ~A()
    {
        //count -=2; /*Keypoint is here. */
                   /*Uncomment it later. But result doesn't change*/
        cout << "destructor called " << id << endl;
    }
};

int A::count = 0;

int main()
{
    A a[2];
    return 0;
}

输出是

constructor called 1
constructor called 2
destructor called 2
destructor called 1

问题是：即使您取消注释，//count -=2; 结果仍然相同。

这是否意味着如果构造函数将静态成员递增 1，那么析构函数也必须将其精确递减 1，并且您不能更改它的行为？

Answer 1

count在调用第一个析构函数后没有任何访问。析构函数完全按照您编写的代码执行，无论是否修改count。count但是除非您以某种方式访问​​，否则您不会看到效果。