Question

我动态创建了两个结构数组（player1和player1temp）。我想使用来自player1的数据进行一些计算并将其保存到player1temp。然后我想将数据从player1temp数据复制到player1。这是一个很好的解决方案吗？

struct team
{
    int value1;
    int value2;
};


int main()
{
    srand(time(NULL));
    team *player1=new team[5];
    for(int i=0; i<5; i++)
    {
        player1[i].value1=rand()%20;
        player1[i].value2=rand()%20;
        cout<<"Player1 index: "<<i<<"     "<<player1[i].value1<<"     "<<player1[i].value2<<"\n";
    }
    team *player1temp=new team[5];
    cout<<"\n\n";
    for(int i=0; i<5; i++)
    {
        player1temp[i].value1=rand()%20;
        player1temp[i].value2=rand()%20;
        cout<<"Player1temp index: "<<i<<"     "<<player1temp[i].value1<<"     "<<player1temp[i].value2<<"\n";
    }
    delete player1;
    player1=player1temp;
    cout<<"\n\n";
    for(int i=0; i<5; i++)
    {
        cout<<"Player1 index: "<<i<<"     "<<player1[i].value1<<"     "<<player1[i].value2<<"\n";
    }
    return 0;
}

Answer 1

两个错误：
如果用于数组，delete应为delete[] 在程序结束之前，应该删除剩余的数组。

矢量会更容易：

struct team
{
    int value1;
    int value2;
};


int main()
{
    srand(time(NULL));

    std::vector<team> player1(5);

    for(int i=0; i<5; i++)
    {
        player1[i].value1=rand()%20;
        player1[i].value2=rand()%20;
        cout<<"Player1 index: "<<i<<"     "<<player1[i].value1<<"     "<<player1[i].value2<<"\n";
    }

    std::vector<team> player1temp = player1;

    cout<<"\n\n";
    for(int i=0; i<5; i++)
    {
        player1temp[i].value1=rand()%20;
        player1temp[i].value2=rand()%20;
        cout<<"Player1temp index: "<<i<<"     "<<player1temp[i].value1<<"     "<<player1temp[i].value2<<"\n";
    }

    player1 = std::move(player1temp);
    //don´t use player1temp anymore now

    cout<<"\n\n";
    for(int i=0; i<5; i++)
    {
        cout<<"Player1 index: "<<i<<"     "<<player1[i].value1<<"     "<<player1[i].value2<<"\n";
    }
    return 0;
}

删除结构数组，更改结构地址

1 个答案: