二元運算符的不良操作數類型％

Question

我正在練習我的java並遇到一些問題。

我想學習從Arraylist中刪除元素，所以我要消除這些可能性。

public static void arrayLists(){ 
     List<Integer> xlist = new ArrayList<Integer>(); 
     for (int x = 0; x < 10; x ++){ 
      xlist.add(x); 
     } 
     for (Iterator<Integer> pointer = xlist.iterator(); pointer.hasNext();){ 
      if (pointer % 2 == 1){ 
       pointer.remove(); 
      } 
     } 
    } 

爲什麼不編譯？ '二元運算符'的錯誤操作數類型'

我認爲問題與列表元素是整數，而我將它們與int（s）進行比較。任何想法如何解決這個問題？

Answer 1

它應該是：

 if (pointer.next() % 2 == 1){ 
      pointer.remove(); 
     } 

指針是Iterator，你不能對其執行％。您必須通過調用pointer.next()來獲取Ierator當前位置處的整數。

Answer 2

替換：

if (pointer % 2 == 1) 

與

if (pointer.next() % 2 == 1) 

Answer 3

有在你的代碼的一些錯誤，工作代碼爲： - 做

public static void arrayLists(){ 
    List<Integer> xlist = new ArrayList<Integer>(); 
    for (int x = 0; x < 10; x++){ //this is not a compiler error but avoid unnecassary spaces , x ++ should be x++ 
     xlist.add(x); 
     } 
    Iterator<Integer> pointer = xlist.iterator(); // write this out of the for loop statement, since we wont be needing it 
    while(pointer.hasNext()){ //its better to use while loop, since there is no increment for counter variable required, the iterator will do that job 
     if (pointer.next() % 2 == 1){ //get current element in iterator by using next() function, pointer.next() 
     pointer.remove(); 
     } 
    } 
} 

Answer 4

變化 - if (pointer.next() % 2 == 1)。 .next實際上將返回對象 公共類的測試{

public static void main(String[] args) { 
     arrayLists(); 
    } 

    public static void arrayLists() { 
     List<Integer> xlist = new ArrayList<Integer>(); 
     for (int x = 0 ; x < 10 ; x++) { 
      xlist.add(x); 
     } 
     for (Iterator<Integer> pointer = xlist.iterator() ; pointer.hasNext() ;) { 
      if (pointer.next() % 2 == 1) { 
       pointer.remove(); 
      } 
     } 

     System.out.println(xlist); 
    } 

} 

輸出

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