C++：默认复制结构函数 执行 浅拷贝

C++, 会默认生成一个复制结构函数, 当类中呈现指针时, 复制会执行浅拷贝, 即只复制指针的地点, 不会复制数据;

所以在类中, 利用指针时, 需要留意; 假如想利用深拷贝, 可以添加复制结构函数.

以下代码, 假如不添加复制结构函数, 则会运行堕落, 但可以通过编译,

运行时, 因为删除(delete[])两次str所指的同一片地点空间, 所以措施无法执行.

代码:

/* 
 * main.cpp 
 * 
 *  Created on: 2014.4.15 
 *      Author: Spike 
 */
      
/*vs2012*/
      
#include <iostream>  
#include <cstring>  
#include <vector>  
#include <memory>  
      
using namespace std;  
      
class CDemo {  
public:  
    CDemo() : str(NULL) {};  
    ~CDemo() {  
        static int i=0;  
        if (str) {  
            std::cout << "&Demo" << i++ << " = " << (int*)this << ", str = " << (int*)str << std::endl;  
            delete[] str;  
        }  
    }  
      
    //复制结构函数  
    CDemo(const CDemo& cd) {  
        this->str = new char[strlen(cd.str) + 1];  
        strcpy(this->str, cd.str);  
    }  
      
    char* str;  
};  
      
int main () {  
    CDemo d1;  
    d1.str = new char[32];  
    strcpy(d1.str, "Caroline");  
    std::vector<CDemo>* a1 = new std::vector<CDemo>();  
    a1->push_back(d1); //执行复制结构函数  
    std::cout << "d1.str = " << d1.str << std::endl;  
    std::cout << "(*a1)[0].str  = " << (*a1)[0].str << std::endl;  
    strcpy(d1.str, "Wendy");  
    std::cout << "d1.str = " << d1.str << std::endl;  
    std::cout << "(*a1)[0].str  = " << (*a1)[0].str << std::endl;  
    delete a1;  
    return 0;  
}

输出:

d1.str = Caroline  
(*a1)[0].str  = Caroline  
d1.str = Wendy  
(*a1)[0].str  = Wendy  
&Demo0 = 0x312570, str = 0x312548  
&Demo1 = 0x22fec8, str = 0x312548

作者：csdn博客 Spike_King