2017-05-14 64 views
0

我想做一个疯狂的库游戏,用户将输入5个变量(名词,形容词,动词,副词,第二名词),然后将在我的子类中使用不同的“库”。问题是,当我输入我的参数时,它将输出为空。我确信有一个简单的方法可以为不同的toStrings使用相同的输入,但是我对新的继承并不确定它的工作原理。子类toString打印null? (使用相同的参数为不同的toStrings)

(我看了看其他的toString空的问题,但没有人跟我的情况相当的工作,或有东西不对它们的构造函数。我相当肯定有什么错我的。)

这里我的短语超:

public class Phrase 
{ 
    private String noun; 
    private String adjective; 
    private String verb; 
    private String adverb; 
    private String noun2; 

    public Phrase (String n, String a, String v, String ad, String n2) 
    { 
     n=noun; 
     a=adjective; 
     v=verb; 
     ad=adverb; 
     n2=noun2; 
    } 

    public String getNoun() 
    {return noun;} 

    public String getAdj() 
    {return adjective;} 

    public String getVerb() 
    {return verb;} 

    public String getAdverb() 
    {return adverb;} 

    public String get2Noun() 
    {return noun2;} 

    //i'll need the get methods in the libs classes and the set methods in the 
    while loop, if user wants to change parameters 

    public void setNoun (String Newnoun) 
    {noun=Newnoun;} 

    public void setAdj (String newAdj) 
    {adjective=newAdj;} 

    public void setVerb (String newVrb) 
    {verb=newVrb;} 

    public void setAdverb (String newAdv) 
    {adverb=newAdv;} 

    public void set2Noun (String newNoun2) 
    {noun2=newNoun2;} 

} 

这里是我的子类:

public class Obama extends Phrase 
{ 
    public Obama(String noun, String adjective, String verb, String adverb, 
      String noun2) 
    {super (noun, adjective, verb, adverb, noun2);} 

    public String getNoun() 
    {return super.getNoun();} 

    public String getAdj() 
    {return super.getAdj();} 

    public String getVerb() 
    {return super.getVerb();} 

    public String getAdverb() 
    {return super.getAdverb();} 

    public String get2Noun (String n) 
    {return super.get2Noun();} 


    public String toString() 
    { 
     return ("there is a " + super.getAdj() + " " + super.getNoun() +" on the 
       floor! It is " + super.getVerb() +"ing " + super.getAdverb() +". Next to it 
       is a " + super.get2Noun()); 
    } 


} 

这里是我的司机:

public class Madlibsdriver 
{ 
    static Scanner scan= new Scanner (System.in); 
    public static void main(String[] args) 
    { 

     System.out.print ("Welcome to Mad libs!"); 
     System.out.println(); 
     System.out.println("-------------------------------------------- "); 
     System.out.println("Enter a noun:"); 
     String ip1= scan.nextLine(); 
     System.out.println("Enter an adjective:"); 
     String ip2= scan.nextLine(); 
     System.out.println ("Enter a verb:"); 
     String ip3= scan.nextLine(); 
     System.out.println ("Enter an adverb (ex: angrily) :"); 
     String ip4= scan.nextLine(); 
     System.out.println ("Enter another noun:"); 
     String ip5= scan.nextLine(); 

     Phrase obama= new Obama(ip1, ip2, ip3, ip4, ip5); 

     System.out.println(); 
     System.out.print (obama.toString()); 

    } 

} 

回答

2

您已经交换了构造函数中的参数,设置了您传入的值(而不是使用它们来设置本地字段)。使用this关键字来捕获此类错误,如

public Phrase (String n, String a, String v, String ad, String n2) 
{ 
    this.noun = n; 
    this.adjective = a; 
    this.verb = v; 
    this.adverb = ad; 
    this.noun2 = n2; 
} 
相关问题