오버로드하는 방법에 문제가 있습니다.

Question

저는 완전히 잘못된 atm을하고 있습니다. im는 코딩에 익숙하지 않습니다. 그냥 시도.

숫자 배열의 평균을 반환하는 두 가지 오버로드 메서드를 만듭니다. 하나의 방법은

public static double average (int[] array) 

다른 방법 헤더 내가이 두 가지를 호출하고 평균을 인쇄 할

public static double average (double[] array) 

내 주요 방법이다. 첫 번째 배열은 {1,2,3,4,5}를 사용하여 정수를 허용하는 평균 메소드를 테스트하고이 배열 {6.0,5.0,4.0,3.0,2.0,1.0}은 매개 변수로 double을 허용하는 평균 메소드를 테스트합니다.

지금 내 코드는 아마도 끔찍한 모양 일 것입니다. 나는 정말로 모른다.

public class methodss 
{ 
    public static void main(String[] args) 
    { 
     //invoke the average int method 
     System.out.println("Average with int: " + average); 
     //invoke the average double method 
     System.out.println("Average with double: " + average); 

     public static double average (int[] array) 
     { 
      int sum = 0, average = 0; 

      array[5] = {1,2,3,4,5} 
      for (int i = 0; i < 10; ++i){ 
       sum+=array[i]; 
       average = sum/5; 
       return average; 
      } 
     } 

     public static double average (double[] array) 
     { 
      int sum = 0, average2 = 0; 

      array[6] = {6.0, 5.0, 4.0, 3.0, 2.0, 1.0}; 
      for (int x = 0; x < 10; ++x){ 
       sum+=array[x]; 
       average = sum/6; 
       return average; 
     } 
    } 
} 

Answer 1

, 대신 시도

// Classes normally are named like proper nouns, so they start with a captial letter 
public class Methods 
{ 
    // main is the entry point of the app. 
    // all the args that are entered on the command line are passed in 
    // as an array of strings. 
    // the keyword static here means that this method belongs to this class 
    // so all invokations are in terms of the class. 
    // i.e. java Methods ... implicitly invokes Methods.main() 
    public static void main(String[] args) 
    { 
     //invoke the average int method 
      System.out.println("Average with int: " + Methods.average(1, 2, 3, 4, 5)); 
     //invoke the average double method 
      System.out.println("Average with double: " + Methods.average(1.0, 5.4, 2.3, 1.6); 
    } 

     /** 
     * takes the average of an arbitrary amount of primitive ints 
     * @param array a varg of ints, so that the caller can specify calls 
     * to the method like average(1) or average(1, 2) or average(new int[10]), 
     * makes calling this function in adhoc ways readable. 
     * @returns a double representing the average of all the elements in the arguments 
     **/ 
     public static double average (int... array) 
     { 
      int sum = 0; 
      double average = 0.0; 
      // loop through the array and stop when the iterator int, i 
      // reaches the amount of elements in the array. 
      // the amount of elements in the array can be read by the 
      // special instance variable on array types called length. 
      for (int i = 0; i < array.length; ++i){ 
       sum += array[i]; 
      } 
      // so that these don't get truncated, inserting a 1.0 to make sure it's 
      // a double. Otherwise the average of (1, 2) would be 1 instead of 1.5 
      average = 1.0 * sum/array.length; 
      return average; 
     } 


     /** 
     * Same as average(int...) but with doubles 
     **/ 
     public static double average (double... array) 
     { 

      double sum = 0.0, average = 0.0; 

      for (int x = 0; x < array.length; ++x){ 
       sum += array[x]; 
      } 
      average = sum/array.length; 
      return average; 
     } 
}