Question

我有一个名为message的数组。当用户输入“下一个”时，我移动到下一个索引。同样，对于以前。我希望能够以一种简单的方式转移到下一个和上一个。我知道下一个但不是之前的简单方法。我该怎么做？

array[] message  = {"Hello", "good", "people", "!"};//index starts at 0
int currentIndex = 0;
int next = 2;//an option which can be used in an if or switch-case
int previous = 1;

转到下一个元素 -

moveNext(){
int currentIndex = (currentIndex + 1) % message.length;
}

//Any neat trick, similar to the one for moveNext() ? 
movePrevious(){
//Yes, you can use an if-else check here, but I am trying to avoid that.
}

修改

Example - Array is of LENGTH = 5, that is indexes 0 to 4. 
When currentIndex = 4 and you enter next, then set currentIndex = 0
When currentIndex = 0 and you enter previous, then set currentIndex = 4

Answer 1

如果添加message.length：

，您仍然可以使用单行余数技术

int currentIndex = (currentIndex + message.length - 1) % message.length;

它对索引0更有帮助，但它似乎是你正在寻找的。请注意添加会引起长度溢出的可能性。

显示逻辑的示例代码：

public class Looper {

    static int len = 4;
    static int currentIndex = 0;

    public static void main(String[] args) {

        System.out.println("Index = " + currentIndex);
        for(int i = 0; i < 5; i++){
            movePrev();
        }

    }// main

    public static void movePrev(){
        currentIndex = (currentIndex + len - 1) % len;
        System.out.println("Moved prev. Index = " + currentIndex);
    }

}// Looper

输出：

指数= 0
  移动了上一个。指数= 3
  移动了上一个。指数= 2
  移动了上一个。指数= 1
  移动了上一个。指数= 0
  移动了上一个。指数= 3

Answer 2

你真正想要的是模数运算，而不是余数运算符（What's the difference?）。这将具有处理负红利所需的语义。幸运的是，从另一个中获取一个并不太难：

public static int Modulus(int dividend, int divisor)
{
    return (dividend % divisor + dividend) % divisor;
}

您现在可以写：

int currentIndex = Modulus(currentIndex + 1, message.length);

或

int currentIndex = Modulus(currentIndex - 1, message.length);

Answer 3

//moveNext()

    if ( Length != 0 && ++currentIndex == Length ) currentIndex = 0;


//movePrevious

    if ( Length != 0 && currentIndex-- == 0 ) currentIndex = Length - 1;

循环数组 - 从最后跳到第一个，反之亦然？

3 个答案: