Compiling error

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  13 years ago

  by bigCountry

  John Wiley

  , United States

  Joined 13 years ago

  Hi, I am having problems when I call getUnion.
  The error message are "getUnion(IntergerSet) in IntergerSet cannot be applied to ()" and "incompatible types"
  where the arrows are.
  Here is my code:

  public void inputSet ()
   {
    Scanner input = new Scanner (System.in);
        
       int value;
       boolean arraySet[] = new boolean[100];
       do
       {  
        System.out.println("\nPlease enter a value or -1 to finish: ");
        value = input.nextInt();
        if(validEntry(value))
         arraySet[value] = true;
        else
         System.out.printf("\nYou enter %d to finish entering values", value, "\n");
       }while(value != -1);
       
       System.out.println("\n\nIf you want to find the union of two sets press 'u': ");
          String unionSets = input.next();
          if("u".equalsIgnoreCase(unionSets))
  --->         System.out.printf("\nUNION", getUnion());
           
      }
     
      //Finding the union of two sets
      public String getUnion(IntergerSet other)
      {
       IntergerSet res = new IntergerSet();
        
       for(int i=0; i<SIZE; i++) 
       {
        res.arraySet[i] = this.arraySet[i]||other.arraySet[i];
       }
  --->     return res;
   }

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• • Reply

  13 years ago

  by Mistuke

  Mistaken Luke

  The Netherlands, Netherlands

  Joined 18 years ago

  There is no overloaded form of getUnion which takes 0 parameters. you only have one which takes 1 parameter. System.out.printf("\nUNION", getUnion()); generates an error because you make a call to getUnion without passing it any argument.

  return res; fails because the type of the method is String and you're trying to return an IntergerSet. change the type of the function to either IntergetSet or make the function return a string. Or overload the toString() method in IntergerSet to make it return what you would view as a string interpretation of the class and then add res.toString() as return.

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