C++语言基础 例程 深复制

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浅复制

#include <iostream>
using namespace std;

class Test
{
private:
    int x;
public:
    Test(int n)  {x=n; }
    Test(const Test& c){x=c.x; }
    void show (){cout<<x<<endl;}
};


int main()
{
    Test a(100);
    Test b(a);
    Test c=a;
    b.show();
    c.show();
    return 0;
}


有问题吗?
#include <iostream>
#include <cstring>
using namespace std;


class Test
{
private:
    int x;
    char *str;
public:
    Test(int n, char *s)
    {
        x=n;
        strcpy(str,s);  //肇事地点,但不是祸端
    }


    Test(const Test& c)
    {
        x=c.x;
        strcpy(str,c.str);
    }
    void show ()
    {
        cout<<x<<","<<str<<endl;
    }
};
int main()
{
    Test a(100,"Hello");
    Test b(a);
    a.show();
    b.show();
    b.show();
    return 0;
}


正解——深复制
#include <iostream>
#include <cstring>
using namespace std;


class Test
{
private:
    int x;
    char *str;
public:
    Test(int n, char *s)
    {
        x=n;
        int m=strlen(s)+1;
        str=new char[m];
        strcpy(str,s);  
    }
    Test(const Test& c)
    {
        x=c.x;
        int m=strlen(c.str);
        str=new char[m];
        strcpy(str,c.str);
    }
    ~Test()
    {
        delete str;
    }
    void show ()
    {
        cout<<x<<","<<str<<endl;
    }
};


int main()
{
    Test a(100,"Hello");
    Test b(a);
    a.show();
    b.show();
    b.show();
    return 0;
}


最危险的修改——貌似对,但一定有机会错
#include <iostream>
#include <cstring>
using namespace std;


class Test
{
private:
    int x;
    char *str;  //指针成员
public:
    Test(int n, char *s){
        x=n;
        str=s;  //不用strcpy(str,s); 
    }
    Test(const Test& c){
        x=c.x;
        str=c.str;
    }
    void show (){        
     cout<<x<<","<<str<<endl;
    }
};


int main()
{
    Test *a;
    a=new Test(100,"Hello");
    Test b(*a);
    a->show();
    b.show();
    delete a;
    b.show();
    return 0;
}


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