私人成員函數一定不能被外部調用嗎？

一般情況下私人成員函數一定不能被外部調用，但是虛擬私人成員函數是可能被外部調用的。請看例子：

 

#include "stdafx.h"<br />#include <iostream><br />using namespace std;</p><p>class base<br />{<br />public:<br />void a() {cout<<"base::a"<<endl;};<br />void b() {cout<<"base::b"<<endl;};</p><p>//以下為虛函數<br />virtual void c() {cout<<"base::c"<<endl;};<br />virtual void d() {cout<<"base::d"<<endl;};<br />virtual void e() {cout<<"base::e"<<endl;};<br />private:<br />virtual void f() {cout<<"base::f"<<endl;};<br />void g() {cout<<"base::g"<<endl;};<br />//私人成員函數<br />};<br />class Derive :public base<br />{<br />public:<br />void a() {cout<<"Derive::a"<<endl;};<br />virtual void b() {cout<<"Derive::b"<<endl;};</p><p>void c() {cout<<"Derive::c"<<endl;};<br />virtual void d() {cout<<"Derive::d"<<endl;};<br />private:<br />void e() {cout<<"Derive::e"<<endl;};<br />//虛函數，可以不加virtual，基類為public時可通過基類訪問<br />public:<br />virtual void f() {cout<<"Derive::f"<<endl;};<br />virtual void g() {cout<<"base::g"<<endl;};<br />};<br />//調用虛函數時其結果為變數初始化類的函數<br />//調用非虛函數時其結果為變數類自身的函數</p><p>int _tmain(int argc, _TCHAR* argv[])<br />{<br /> base a;<br />Derive *b=(Derive *)&a;</p><p>b->a(); //Derive::a<br />//b->b(); //b是虛擬表裡的函數，轉化後不存在<br /> b->c(); //base::c<br />b->d(); //base::d<br />//b->e();<br />b->f(); //base::f　　//私人成員函數被調用<br />//b->g(); //出錯。</p><p>Derive d;<br />base *c=(base *)&d;</p><p>c->a(); //base::a<br />c->b(); //base::b<br />c->c(); //Derive::c<br />c->d(); //Derive::d<br />c->e(); //Derive::e<br />//c->f(); //private<br />//c->g(); //private </p><p>int ch;<br />cin>>ch;</p><p>return 0;<br />}<br />

 

想要深刻理解以上代碼，其核心是要理解虛函數表和記憶體分布，推薦文章：
陳皓專欄：http://blog.csdn.net/haoel/archive/2007/12/18/1948051.aspx
小寶--讀書筆記：http://blog.chinaunix.net/u/16292/showart_676270.html