我如何克服此代码中的 java.lang.ArrayIndexOutOfBoundsException: 2 ？
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How do I overcome java.lang.ArrayIndexOutOfBoundsException: 2 in this code?
提问by user1215225
When I tested this code, it gave me java.lang.ArrayIndexOutOfBoundsException: 2 The code isn't finished yet but this is what I am trying to do: A program that asks the user for two numbers. Print all numbers from the smaller number to the larger number (inclusive) in a random order. Using this order, find the largest sum of 2 numbers adjacent to one another. Could some please point out to me what needs to be fixed as well as how to fix it? Thanks.
当我测试这段代码时，它给了我 java.lang.ArrayIndexOutOfBoundsException: 2 代码还没有完成，但这就是我想要做的：一个要求用户输入两个数字的程序。以随机顺序打印从小数到大数（含）的所有数字。使用此顺序，找出彼此相邻的 2 个数字的最大和。能否请一些人向我指出需要修复的内容以及如何修复它？谢谢。
boolean isNum = false;
int in1 = 0;
int in2 = 0;
int size = 0;
int largestsum = 0;
while (!isNum) {
try {
System.out.print("Enter a number: ");
in1 = Integer.parseInt(in.readLine());
boolean isSame = false;
while (!isSame) {
System.out.print("Enter a number: ");
try {
in2 = Integer.parseInt(in.readLine());
if (in1 != in2) {
isSame = true;
} else {
System.out.println("Same number.");
}
} catch (NumberFormatException nfe) {
System.out.println("Not a number.");
}
}
isNum = true;
} catch (NumberFormatException nfe) { //catches error
System.out.println("Not a number."); //informs user of the error
}
}
if (in1 > in2) {
size = in1 - in2 + 1;
int[] num = new int[size];
for (int i = 0; i < size; i++) {
num[i] = in2 + i;
}
int[] order = new int[size];
boolean[] used = new boolean[size];
for (int i = 0; i < size; i++) {
int r = (int) ((size) * (Math.random()));
while (used[r]) {
r = (int) ((size) * Math.random());
}
order[i] = r; //fill array
used[r] = true; //r is now used
}
for (int i = 0; i < size; i++) {
if (num[order[i]] + num[order[i + 1]] > largestsum) {
largestsum = num[order[i]] + num[order[i + 1]];
}
}
采纳答案by wattostudios
The last forloop will overflow when i=size-1
当 i=size-1 时，最后一个for循环将溢出
In your question, you want to get the sum between 2 adjacent numbers. Your logic in the last forloop says that you calculate the sum by adding number ito i+1. The last item in your array is at position size-1, so when you then try to add this to the next adjacent number, (size-1)+1overflows the array.
在您的问题中，您想获得两个相邻数字之间的总和。您在最后一个for循环中的逻辑表示您通过将数字i添加到i+1来计算总和。数组中的最后一项位于size-1位置，因此当您尝试将其添加到下一个相邻数字时，(size-1)+1 会溢出数组。
To fix the problem, your last for loop should only continue until i<size-1, like so...
为了解决这个问题，你的最后一个 for 循环应该只持续到i<size-1，就像这样......
for (int i = 0; i < size-1; i++) {
if (num[order[i]] + num[order[i + 1]] > largestsum) {
largestsum = num[order[i]] + num[order[i + 1]];
}
}
With this fixed logic, the largest item in the forloop is size-2, and when you add 1 to it (for the next adjacent number), it will choose item (size-2)+1, which is the last item in the array, rather than overflowing the array.
使用这个固定逻辑，for循环中最大的项目是size-2，当你加 1 时（对于下一个相邻的数字），它会选择项目(size-2)+1，这是最后一个项目数组，而不是溢出数组。
回答by VeeArr
My guess is the problem is in your very last for loop. Since ordercan contain values ranging from 0to size, num[order[i]]is always valid, but in the case where order[i]==size, num[order[i+1]]is out of bounds.
我的猜测是问题出在您的最后一个 for 循环中。由于order可以包含范围从0到 的值size，num[order[i]]因此始终有效，但在 的情况下，order[i]==size,num[order[i+1]]超出范围。
