C++ 构造和析构的顺序

构造函数调用顺序

1. 调用虚基类构造函数（从左到右）
2. 调用非虚基类构造函数（从左到右）
3. 调用成员变量构造函数（按声明顺序从前往后，而非初始化列表顺序）
4. 调用类自身构造函数的语句

析构函数调用顺序

与构造函数相反

1. 执行自身析构函数语句
2. 析构成员变量（从后往前）
3. 调用非虚基类构造函数（从右到左）
4. 调用虚基类构造函数（从右到左）

例子

#include <bits/stdc++.h>
using namespace std;

class Base1 {
public:
    Base1() {
        cout << "Base1 ctor" << endl;
    }

    ~Base1() {
        cout << "Base1 dtor" << endl;
    }
};

class Base2 {
public:
    Base2() {
        cout << "Base2 ctor" << endl;
    }

    ~Base2() {
        cout << "Base2 dtor" << endl;
    }
};

class VBase {
public:
    VBase() {
        cout << "VBase ctor" << endl;
    }

    ~VBase() {
        cout << "VBase dtor" << endl;
    }
};

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

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

class Derived: public Base1, public Base2, virtual public VBase {
private:
    A a;
    B b;

public:
    Derived(): b(), a() {
        cout << "Derived ctor" << endl;
    }
    ~Derived() {
        cout << "Derived dtor" << endl;
    }
};

int main(void) {
    Derived der;
    // Base1 *base = new Derived;
    // delete base;
    return 0;
}

VBase ctor
Base1 ctor
Base2 ctor
A ctor
B ctor
Derived ctor
Derived dtor
B dtor
A dtor
Base2 dtor
Base1 dtor
VBase dtor

完全符合规则

多态的基类析构函数要声明为 virtual

下面如果使用 Base1 *base = new Derived; delete base; 呢

VBase ctor
Base1 ctor
Base2 ctor
A ctor
B ctor
Derived ctor
Base1 dtor

思考哪出了问题？

其实是为了实现多态，则必须声明基类的析构函数为 virtual，本题可以将 Base1 的析构函数声明为 virtual

结果：

VBase ctor
Base1 ctor
Base2 ctor
A ctor
B ctor
Derived ctor
Derived dtor
B dtor
A dtor
Base2 dtor
Base1 dtor
VBase dtor