2014-04-15 63 views
0

所以,我是超新的Java,但我正在尝试制作一个特定于类的程序,以便我可以在Java中进行游戏。这里是我的代码:如何正确使用Java方法?

import java.util.Scanner; 
public class boxtype { 
public static void main(String args[]) { 
    String[] melee = {"Crowbar", "Bowie Knife", "Butterfly Knife", "Knuckleduster"}; 
    String[] pistol = {"Colt .22", "Magnum .45", "P250", "9mm Pistol"}; 
    String[] assault = {"AK47", "M4A1", "M16", "SMG", "Mac10", "Minigun (HGE)"}; 
    String[] shotgunsniper = {"Shotgun", "Benelli S90", "Sniper Rifle"}; 
    String[] attachments = {"Laser Sight", "Silencer", "Scope", "Auto-target"}; 
    Scanner scan = new Scanner(System.in); 
    String xy = scan.nextLine(); 
    if (xy.equals("spyclass")) { 
     spyClass(); 
    } 

} 
private static void spyClass(String[] assault, String[] attachments, String[] pistol) { 
    // TODO Auto-generated method stub 
    System.out.println("Spy class: "); 
    System.out.println("Primary weapon: " + assault[2] + " + " + attachments[2]); 
    System.out.println("Secondary weapon: " + pistol[1] + " + " + attachments[2]); 
    System.out.println(""); 
} 

} 

基本上会发生什么,就是Eclipse返回一个错误,指出“spyClass不适用”。我仍在研究如何解决,但是。

呼叫
+0

你有太多的花括号。除去'spyClass'方法之上的一个。 – Makoto

回答

2

到spyClass你没有通过parameeters

它应该是:

if (xy.equals("spyclass")) { 
    spyClass(assault, attachments, pistol); 
} 
+0

非常感谢你,我找到了一种方法_kind of_ works - 将变量放在每个方法中。感谢这个想法,我会使用它。 – Declan

0

spyClass方法需要一堆参数,但是你不给它任何。上面写着

spyClass(); 

行也许应该像

spyClass(assault, attachments, piston); 
0

spyClass可能不应该static,你需要的时候你在main方法调用它的参数传递给它。 spyClass(assault, attachments, pistol);