#include<iostream>using namespace std;//繼承與組合混搭情況下，構(gòu)造和析構(gòu)調(diào)用原則 class Grandfather { public:Grandfather( char * g){this->G_t1=g;cout<<"G_t1====== "<<G_t1;cout<<"我是頂頭的爺爺?shù)臉?gòu)造函數(shù)\n";}~Grandfather(){cout<<"我是頂頭的爺爺?shù)奈鰳?gòu)函數(shù)\n";} protected:char *G_t1; private: }; class Father:public Grandfather { public:Father( char *f):Grandfather("hahahah"){F_t1=f;cout<<"F_t1====== "<<F_t1;cout<<"這是父親的構(gòu)造函數(shù)\n";}~Father(){cout<<"這是父親的析構(gòu)函數(shù) \n";} protected:char *F_t1; private: }; class Child:public Father { public:Child():Father("dsa"),g1("sfsdfds"),g2("g2")//不懂{cout<<"這是孩子的構(gòu)造函數(shù) \n";}~Child(){cout<<"這是孩子的析構(gòu)函數(shù) \n";} protected:int C_t1;Grandfather g1;//這里這兩句有疑問(wèn)，為什么Grandfather明明必須有形參，但是為什么這里初始化參數(shù)反而不對(duì)了呢？Grandfather g2; private: };void objplay() {Child c1; } int main() {objplay();system("pause");return 0; }

　　不明白的是這兩句

?但是這好像是套路，必須這么做呢

就像例如：

#include<iostream>using namespace std;//繼承與組合混搭情況下，構(gòu)造和析構(gòu)調(diào)用原則 class Grandfather { public:Grandfather( char * g){this->G_t1=g;cout<<"G_t1====== "<<G_t1;cout<<"我是頂頭的爺爺?shù)臉?gòu)造函數(shù)\n";}~Grandfather(){cout<<"我是頂頭的爺爺?shù)奈鰳?gòu)函數(shù)\n";} protected:char *G_t1; private: }; class Father:public Grandfather { public:Father( char *f):Grandfather("hahahah"){F_t1=f;cout<<"F_t1====== "<<F_t1;cout<<"這是父親的構(gòu)造函數(shù)\n";}~Father(){cout<<"這是父親的析構(gòu)函數(shù) \n";} protected:char *F_t1; private: }; class Child:public Father { public:Child():Father("dsa"),g1("sfsdfds"),g2("g2"),f1("f1")//不懂{cout<<"這是孩子的構(gòu)造函數(shù) \n";}~Child(){cout<<"這是孩子的析構(gòu)函數(shù) \n";} protected:int C_t1;Father f1;Grandfather g1;//這里這兩句有疑問(wèn)，為什么Grandfather明明必須有形參，但是為什么這里初始化參數(shù)反而不對(duì)了呢？Grandfather g2; private: };void objplay() {Child c1; } int main() {objplay();system("pause");return 0; }

　　運(yùn)行結(jié)果是：

?

發(fā)現(xiàn)一個(gè)真理沒(méi)？只要調(diào)用一個(gè)Father就必須調(diào)用一個(gè)爺爺啊

套路

轉(zhuǎn)載于:https://www.cnblogs.com/xiaochige/p/6613022.html

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