一、實驗?zāi)康?．理解多重派生的定義；2．理解多重派生中構(gòu)造函數(shù)與析構(gòu)函數(shù)的調(diào)用順序；3．理解多重派生中虛擬基類的作用；二、實驗內(nèi)容2.1練習(xí)（一）：1．理解下面的程序，并在VC++6.0下運行查看結(jié)果，回答程序后面的問題。class CBase1{public:CBase1(int a)? :a(a){? cout<<"base1 structure..."<<endl;}~CBase1(){? cout<<"base1 destructure..."<<endl;}void print(){? cout<<"a="<<a<<endl;}protected:int a;};class CBase2{public:CBase2(int b)? :b(b){? cout<<"base2 structure..."<<endl;}~CBase2(){? cout<<"base2 destructure..."<<endl;}void print(){? cout<<"b="<<b<<endl;}protected:int b;};class CDerive : public CBase1, public CBase2{public:CDerive(){? cout<<"derive structure..."<<endl;}~CDerive(){? cout<<"derive destructure..."<<endl;}void print(){? CBase1::print();? CBase2::print();? b1.print();? b2.print();? cout<<"c="<<c<<endl;}private:CBase1 b1;CBase2 b2;int c;};void main(){CDerive d;d.print();}問題一：改正以上程序中存在的錯誤，并分析該程序的輸出結(jié)果。2．理解下面的程序，并在VC++6.0下運行查看結(jié)果，回答程序后面的問題。#include "iostream.h"class CBase{public:CBase(int a)? :a(a){}int a;};class CDerive1 : public CBase{public:CDerive1(int a)? :CBase(a){}};class CDerive2 : public CBase{public:CDerive2(int a)? :CBase(a){}};class CDerive : public CDerive1,public CDerive2{public:CDerive(int a,int b)? :CDerive1(a),CDerive2(b){}};void main(){CDerive d(1,2);cout<<d.a<<endl;}問題一：在不改變原有程序意圖的前提下，改正以上程序，并使程序正確輸出。