C++重载，重写，重定义的区别

这三个概念一直比较混乱，正好看到了篇帖子，感觉不错，梳理了一下，详见下图：

参考文章：http://blog.sina.com.cn/s/blog_8ddc5c2e01013hbd.html

测试代码：

test.h

#ifndef _TEST_H
#define _TEST_H

#include 
using std::cout;
using std::endl;

class Base
{
public:
    virtual void func1(int a)
    {
       cout << "base func1 int :" << a << endl; 
    }

    void func2(int a)
    {
       cout << "base func2 int :" << a << endl; 
    }
};

class Son: public Base
{
public:
    /************ overload ***********/
    void func3(std::string str)
    {
       cout << "overload func3 str :" << str << endl; 
    }
    void func3(int a, int b)
    {
       cout << "overload func3 int a, int b:" << a << b << endl; 
    }

    /************ override ***********/
    virtual void func1(int a)
    {
       cout << "son override func1 int :" << a << endl; 
    }

    /************ redifine ***********/
    void func2(char *str)
    {
       cout << "son redifine func2 str :" << str << endl; 
    }

    void func2(char ch)
    {
       cout << "son redifine func2 char :" << ch << endl; 
    }

};
#endif

main.cpp

#include
#include "test.h"
using namespace std;

int main()
{
    Son *son = new Son();
    cout << " ************ overload ***********" << endl;
    son->func3("hello world!");
    son->func3("10, 20!");

    cout << " ************ override ***********" << endl;
    Base *base = new Base();
    base->func1(2);
    son->func1(1);

    base = son;
    base->func1(3);
    son->Base::func1(4);

    cout << " ************ redifine ***********" << endl;
    son->Base::func2(5);
    char str[] = "hello fox!";
    son->func2(str);
    son->func2('h');

    return 0;
}

编译：

g++ main.cpp -I. --std=c++11

输出：

 ************ overload ***********
overload func3 str :hello world!
overload func3 str :10, 20!
 ************ override ***********
base func1 int :2
son override func1 int :1
son override func1 int :3
base func1 int :4
 ************ redifine ***********
base func2 int :5
son redifine func2 str :hello fox!
son redifine func2 char :h

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