從數組列表

Question

計算平均我得到

sum = sum + arr[i]; // "i cannot be resolved to a variable" 

爲什麼是一個錯誤？
「我」是早些時候宣佈，我嘗試使用值arr[i]找到並從文本文件一個ArrayList的平均稱爲numbers.txt

public class Statistics { 

    public static void main(String[] args) { 

     int [] numbers = readFiles("./bin/q1/numbers.txt"); 
     System.out.println(Arrays.toString(numbers)); 

    } 

    private static int[] readFiles(String file) { 

     try { 
      File f = new File(file); 
      Scanner s = new Scanner(f); 
      int ctr = 0; 

      //counting number of integers in numbers.txt 
      while (s.hasNextInt()) { 
       ctr++; 
       s.nextInt(); 
      } 

      int [] arr = new int[ctr]; 
      Scanner s1 = new Scanner(f); 

      //reads integers from file and assigns it to array 
      for (int i = 0; i < arr.length; i ++) 
       arr[i] = s1.nextInt(); 

      int sum = 0; 
      sum = sum + arr[i]; 
      float average = sum/(float) arr.length; 
      System.out.println("average of array is: " + average + "\n"); 
      System.out.println("length of array is: " + arr.length + " integers \n"); 
      return arr; 

     } 

     catch (Exception e) { 
      return null; 
     } 
    } 
}; 

Answer 1

你遇到的問題，在i你宣佈只存在於以前for-loop

//reads integers from file and assigns it to array 
for (int i = 0; i < arr.length; i ++) 
    arr[i] = s1.nextInt(); 

的i將for-loop後銷燬，這意味着你不能訪問i了，你可能會發現更多的德在here

尾巴解決你的問題，你有很多選擇，比如：

創造了另一個for-loop

//reads integers from file and assigns it to array 
for (int i = 0; i < arr.length; i ++) 
    arr[i] = s1.nextInt(); 

int sum = 0; 
for (int i = 0; i < arr.length; i ++) 
    sum = sum + arr[i]; 

float average = sum/(float) arr.length; 

包含了過去for-loop

//reads integers from file and assigns it to array 
int sum = 0; 
for (int i = 0; i < arr.length; i ++) { 
    arr[i] = s1.nextInt(); 
    sum = sum + arr[i]; 
} 

float average = sum/(float) arr.length;