C++在多继承时构造函数调用的顺序

今天看书时注意到C++在多继承时，基类的构造顺序是按照继承时给定顺序执行的，析构的顺序正好相反！而与它们初始化的顺序没有任何关系，下面随便写了一个Demo程序来验证C++在多继承时基类的构造顺序。

#include <iostream>
#include <cstdlib>

using namespace std;

class a
{
      public:
             a(){cout<<"a constructor!!"<<endl;}
             ~a(){cout<<"a deconstructor!!"<<endl;}      
};

class b
{
      public:
             b(){cout<<"b constructor!!"<<endl;}
             ~b(){cout<<"b deconstructor!!"<<endl;}    
};

class c
{
      public:
             c(){cout<<"c constructor!!"<<endl;}
             ~c(){cout<<"c deconstructor!!"<<endl;}    
};

class d
{
      public:
             d(){cout<<"d constructor!!"<<endl;}
             ~d(){cout<<"d deconstructor!!"<<endl;}    
};

class e
{
      public:
             e(){cout<<"e constructor!!"<<endl;}
             ~e(){cout<<"e deconstructor!!"<<endl;}    
};

class f:public e,public b,public c,public d,public a
{
      public:
             f(){cout<<"f constructor!!"<<endl;}
             ~f(){cout<<"f deconstructor!!"<<endl;}    
};


int main(int argc,char * args[])
{
    f testObj;
    
    system("pause");    
    return EXIT_SUCCESS;
}

类f分别继承自类e,类b,类c,类d和类a，在main函数中定义了一个testObj对象，这个对象的构造顺序是e->b->c->d->a，析构顺序正好相反。


程序运行结果如下图所示