Question

很抱歉这是代码有点长，但我需要得到正确的方案。

为什么此代码输出所有'C'？

import java.util.Hashtable;

public class Main {  
    public static ContainsTheHash containsthehash = new ContainsTheHash();  
    public static StoresValues storesvalues = new StoresValues();  
    public static GetsValuesAndPrints getsvaluesandprints = new GetsValuesAndPrints();  
    public static void main(String[] args) {}  

}

class ContainsTheHash {

    Hashtable script_code = new Hashtable();  
    public Contains_The_Hash() {};  
    public void put(long key, Script_Hash_Type sht){script_code.put(key, sht);}  
    public ScriptHashType get(long key){return (Script_Hash_Type)script_code.get(key);}  

}

class ScriptHashType {

     String string;  
     public ScriptHashType(){}  
     public String getstring () {return string;}  
     public void setstring(String str){string = str;}  

}



 class StoresValues {

     public StoresValues(){
         put();
     }
     public void put(){


          ScriptHashType sht = new ScriptHashType();  
          sht.setstring("A");  
          Main.contains_the_hash.put(1,sht);  
          sht.setstring("B");  
          Main.contains_the_hash.put(2,sht);  
          sht.setstring("C");  
          Main.contains_the_hash.put(3,sht);  
     }  

}

class GetsValuesAndPrints {

    public GetsValuesAndPrints(){  

           //should print "A\n B\n  C\n"  
           long temp = 1;  
           System.out.println(get(temp));  
           temp = 2;  
           System.out.println(get(temp));  
           temp = 3;  
           System.out.println(get(temp));  
    };


    public String get(long key){  

        return new String(((Script_Hash_Type)Main.contains_the_hash.get(key)).getstring());  

   }
}

Answer 1

更改：

ScriptHashType sht = new ScriptHashType();
sht.setstring("A");
Main.contains_the_hash.put(1,sht);
sht.setstring("B");
Main.contains_the_hash.put(2,sht);
sht.setstring("C");
Main.contains_the_hash.put(3,sht);

到

ScriptHashType sht = new ScriptHashType();
sht.setstring("A");
Main.contains_the_hash.put(1,sht);
sht = new ScriptHashType();
sht.setstring("B");
Main.contains_the_hash.put(2,sht);
sht = new ScriptHashType();
sht.setstring("C");
Main.contains_the_hash.put(3,sht);

在第一段代码中，每次更新同一个对象

Answer 2

class StoresValues {

     public StoresValues() {
         put();
     };

     public void put() {
          ScriptHashType sht = new ScriptHashType();  
          sht.setstring("A");  
          Main.contains_the_hash.put(1,sht);  
          sht.setstring("B");  
          Main.contains_the_hash.put(2,sht);  
          sht.setstring("C");  
          Main.contains_the_hash.put(3,sht);  
     }

使用new运算符只构建一个单个对象sht。您添加该对象3次。该对象上的最后一个设置字符串将字符串设置为“C”。因为只有一个对象sht，单个对象将具有值“C”。

你应该做的是：

 public void put() {
  ScriptHashType sht = new ScriptHashType();  
  sht.setstring("A");  
  Main.contains_the_hash.put(1,sht);  
  sht = new ScriptHashType();  
  sht.setstring("B");  
  Main.contains_the_hash.put(2,sht);  
  sht = new ScriptHashType();  
  sht.setstring("C");  
  Main.contains_the_hash.put(3,sht);  
 }

Answer 3

您只是将一个ScriptHashType实例存储到Hashtable中，然后操纵其值。要获得您期望的行为，您需要创建三个ScriptHashType实例，每个实例用于一个所需的值。

java中的Hashtable给我最后一个存储的值，但不是正确的值

3 个答案: