Question

我需要在main（）

中创建一个对象Pokemon

将它分配到PokemonWorld类中，并让PokemonWolrd决定哪个PokemonStation是这个宠物小精灵需要去的

我厌倦了分开获取数据（得到名字和hp）并聚在一起（得到一个口袋妖怪类）

但都失败了

#include <iostream>
#include <cstring>
#include <iomanip>

using namespace std;



class Pokemon {
public:
    Pokemon() {};

    Pokemon(char x[], int n) {
        strncpy_s(name, x, 10);
        hp = n;
    };
private:
    char name[10];
    int hp;
};

class PokemonStation {
private:
    Pokemon **list= new Pokemon*[1000];
public:
    PokemonStation() {};
    PokemonStation(int x) {
        id = x;
    };
    int id;
    void assigntoList(int i,Pokemon x)
    { 
        if (i > 0)
            i--;
        list[i] = new Pokemon(x);
        cout << "creat" << list[i];
    };
};

class PokemonWorld {
private:
    char name[10];
public:
    PokemonStation s1;
    PokemonStation s2;
    PokemonWorld() {};
    PokemonWorld(char x[], int y=1, int z=2) {
        strncpy_s(name, x, 10);
        PokemonStation s1(y);
        PokemonStation s2(z);

    };
    const char* const getName() const{
        return name;
    };
    void assigntoStation(int i,Pokemon x) {
        if (i == 0 || i % 2 == 0)
            s1.assigntoList(i, x);
        else
            s2.assigntoList(i, x);
    };
};

void main() {
    int number,hp,i;
    char name[10];
    cout << "What is the World Name ?" <<endl;
    cin >> name;

    PokemonWorld world(name);

    cout << "Please input the number of Pokemon in the " << world.getName() <<" world:" << endl;
    cin >> number;
    Pokemon **mon = new Pokemon*[number];
    cout << "Please input the characteristics of all Pokemon: Name HP" << endl;
    for (i = 0;i < number;i++)
    {
        cin >> name >> hp;
        mon[i] = new Pokemon(name, hp);
        world.assigntoStation(i,*(mon[i]));
    }
    for (i = 0;i < number;i++)

    cout << "world is " << world.getName() << endl;


    system("pause");

};

Answer 1

在C ++中，您应该将std::vectors用于动态事物列表，并使用std::strings用于文本。如果你了解Java，那就像ArrayList和String。（要使用这些，请确保#include <vector>和<string>。）

例如，您的Pokemon课程已被name重写为string：

class Pokemon {
public:
    Pokemon() {}

    Pokemon(string name, int hp):name(name), hp(hp) { // construct the fields directly

    }
private:
    string name;
    int hp;
};

您的PokemonStation课程，使用vector为口袋妖怪列表重写：

class PokemonStation {
private:
    vector<Pokemon> list;
public:
    PokemonStation() {}
    PokemonStation(int x):id(x) {}
    int id;
    void assignToList(Pokemon x)
    { 
        list.push_back(x); // add x to the list
    }
};

如果您要使用Pokemon打印cout <<，那么您必须重载<<运算符以定义要打印的内容。将其添加到您的班级中：

class Pokemon {
public:
    ...
    friend ostream& operator<<(ostream& out, const Pokemon& p) {
        out << p.name; // just print the name
        return out;
    }
    ...
};

只需确保您cout Pokemon Pokemon，而不是指向Pokemon*（WHERE .... AND CONTAINS( jsondata, 'NEAR(('Payments,460),1)')），然后您就赢了。得到一个地址。

如何将类对象分配到类中

1 个答案: