如何修复数组索引超出范围的错误? [英] How to fix array index out of bounds error?
本文介绍了如何修复数组索引超出范围的错误?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我得到的错误
Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 610
at Fib.sorted(Fib.java:67)
at Fib.main(Fib.java:17)
我的代码
public class Fib
{
public static void main(String args[])
{
System.out.println(Arrays.toString( fiblist) );
System.out.println(Fib.add());
System.out.println(Fib.square());
System.out.println(Fib.reversal());
System.out.println(Fib.sorted());
}
public static int fiblist[] = {1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765};
public static int fiblen = fiblist.length;
public Fib()
{
// Do nothing
}
public static ArrayList<Integer> sorted()
{
ArrayList sorted = new ArrayList();
for(int counter = 0; counter < fiblist[4]; counter++ )
{
int temp1 = fiblist[counter];
System.out.println("Elements stored " + temp1);
}
for(int counter = fiblist[14]; counter < fiblist[19]; counter++)
{
int temp2 = fiblist[counter];
System.out.println("Last Elements stored " + temp2);
}
return sorted;
}
}
我正在尝试将数组的最后5个元素存储在temp 2中.然后,我将其切换.有没有更简单的方法可以做到这一点?切换数组的前五个元素与后五个元素?您将如何通过for循环切换它们?
I'm trying to store the last 5 elements of my array in temp 2. Then I will switch them. Is there an easier way to do this? Switch the first five elements of an array with the last five? How would you switch them with a for loop?
推荐答案
有效
for(int i=0;i<fiblist.length;i++){
System.out.print(fiblist[i]+",");
}
System.out.println();
for (int i=0;i<5;i++){
temp=fiblist[i];
fiblist[i]=fiblist[fiblist.length-i-1];
//the first ellement= the last
//the second=second from last...
fiblist[fiblist.length-1-i]=temp;
}
for(int i=0;i<fiblist.length;i++){
System.out.print(fiblist[i]+",");
}
输出:
1,1,2,3,5,8,13,21,34,55,89,144,233,377,610,987,1597,2584,4181,6765,
6765,4181,2584,1597,987,8,13,21,34,55,89,144,233,377,610,5,3,2,1,1,
这篇关于如何修复数组索引超出范围的错误?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
查看全文