Question

即时创建一个口袋妖怪对战游戏，我根据用户输入在if语句中实例化pokemon对象。这显然是一个问题，因为我不能在if语句之外使用这些对象，但必须在每个用户上创建不同的pokemon对象。我需要在if语句之外使用这些对象及其方法的一般方法，或者我需要另一种方法来创建不同于用户输入的对象。请记住，我是java的初学者，谢谢，我需要帮助，任何人的帮助将不胜感激。

代码很长，但这里有一段。

   if(answers == 1) {
      Gible pokemon1 = new Gible("Gible","Gabite & Garchomp", "Achieving Level 24","Dragon Type","Ground Type", "Slash, Tackle, Sandtomb, Dragonclaw", "Monster & Dragon", true , "Gabite & Garchomp", "Slash!", "Tackle!" , "Sandtomb!", "Dragonclaw!", 239, 189, 179, 189, 183, 320);
      System.out.println(pokemon1);
     }
     else {
      Squirtle pokemon2 = new Squirtle("Squirtle","Wartortle & Blastoise", "Achieving Level 16","Water", "None", "Tackle, Water Gun, Hydro Pump, Skull Bash", "Monster & Water 1", true , "Wartortle & Blastoise", "Tackle!", "Water Gun!" , "Hydro Pump!", "Skull Bash!", 195, 229, 199, 227, 185, 292);
      System.out.println(pokemon2);  

       }

if语句只是用户输入决定他们想要哪个口袋妖怪（＃1或＃2）。我创建的对象直接绑定到抽象类的子类，该类返回一堆stats和info（基本上我放入参数中的东西）。

Answer 1

您可以在if语句之外声明对象，然后将其初始化。所以不要这样：

if (some test) {
    PokemonObject po = new PokemonObject();
    // ... use po
} else if (another test) {
    PokemonObject po = new PokemonObject();
    // ... use po
} else if (a third test) {
    PokemonObject po = new PokemonObject();
    // ... use po
}
// try to use po (doesn't compile)

请改用：

PokemonObject po = null;
if (some test) {
    po = new PokemonObject();
    // ... use po
} else if (another test) {
    po = new PokemonObject();
    // ... use po
} else if (a third test) {
    po = new PokemonObject();
    // ... use po
}
if (po != null) {
    // try to use po (works!)
}

编辑：如果你有专门的pokemon对象类，那么你可以定义一个基类（类可以扩展）或一个接口（类可以实现），定义所有pokemon对象共有的行为。您的代码可能如下所示：

PokemonObject po = null; // base class type for Gibble and Squirtle
if (answers == 1) {
    Gible pokemon = new Gible("Gible","Gabite & Garchomp", "Achieving Level 24","Dragon Type","Ground Type", "Slash, Tackle, Sandtomb, Dragonclaw", "Monster & Dragon", true , "Gabite & Garchomp", "Slash!", "Tackle!" , "Sandtomb!", "Dragonclaw!", 239, 189, 179, 189, 183, 320);
    // do Gible-specific stuff here
    po = pokemon;
} else {
    Squirtle pokemon = new Squirtle("Squirtle","Wartortle & Blastoise", "Achieving Level 16","Water", "None", "Tackle, Water Gun, Hydro Pump, Skull Bash", "Monster & Water 1", true , "Wartortle & Blastoise", "Tackle!", "Water Gun!" , "Hydro Pump!", "Skull Bash!", 195, 229, 199, 227, 185, 292);
    // do Squirtle-specific stuff here
    po = pokemon;
}
// do general Pokemon stuff here
System.out.println(pokemon.getName());

如果使用基类而不是接口，基类构造函数（或构造函数）将只接受对所有Pokemon对象有意义的参数。它也只会定义所有Pokemon对象共有的操作（如getName()）。

另外，构造函数调用看起来非常复杂且容易出错。您可能希望使用builder pattern或相关技术来避免使用too many parameters的构造函数。

Answer 2

试试这个：

public Gible pokemon1;
public Squirtle pokemon2;
    if(answers == 1) {
          this.pokemon1 = new Gible("Gible","Gabite & Garchomp", "Achieving Level 24","Dragon Type","Ground Type", "Slash, Tackle, Sandtomb, Dragonclaw", "Monster & Dragon", true , "Gabite & Garchomp", "Slash!", "Tackle!" , "Sandtomb!", "Dragonclaw!", 239, 189, 179, 189, 183, 320);
          System.out.println(pokemon1);
         }
         else {
          this.pokemon2 = new Squirtle("Squirtle","Wartortle & Blastoise", "Achieving Level 16","Water", "None", "Tackle, Water Gun, Hydro Pump, Skull Bash", "Monster & Water 1", true , "Wartortle & Blastoise", "Tackle!", "Water Gun!" , "Hydro Pump!", "Skull Bash!", 195, 229, 199, 227, 185, 292);
          System.out.println(pokemon2);  

           }

现在你可以从每个地方调用这两个变量。

Answer 3

由于Ted Hopp建议而不是在IF块内部进行声明和实例化，您可以声明IF块之外的对象，当您进入IF块时，您可以实例化对象。

GibleClass gb = null;
SquirtleClass sq = null;

        if(answer == 1){
        gb = new Gible("Gible","Gabite & Garchomp", "Achieving Level 24","Dragon Type","Ground Type", "Slash, Tackle, Sandtomb, Dragonclaw", "Monster & Dragon", true , "Gabite & Garchomp", "Slash!", "Tackle!" , "Sandtomb!", "Dragonclaw!", 239, 189, 179, 189, 183, 320);
        System.out.println(gb);

        } else {
        sq = new Squirtle("Squirtle","Wartortle & Blastoise", "Achieving Level 16","Water", "None", "Tackle, Water Gun, Hydro Pump, Skull Bash", "Monster & Water 1", true , "Wartortle & Blastoise", "Tackle!", "Water Gun!" , "Hydro Pump!", "Skull Bash!", 195, 229, 199, 227, 185, 292); 
        System.out.println(sq);
        }

如何在其范围之外使用对象

3 个答案: