将正确的代码放在一起时出现Java ArrayIndexOutOfBoundsException错误[重复]

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How to avoid ArrayIndexOutOfBoundsException or IndexOutOfBoundsException? [duplicate] 2 answers

如何避免ArrayIndexOutOfBoundsException或IndexOutOfBoundsException? [重复] 2个答案

I have a few string problems that I need to put together for a complete homework assignment. They all work correctly by themselves, but when I put them together in the main function, the last one that finds the smallest word in a string gives an error. Anyone know why?

我有一些字符串问题,我需要将它们放在一起完成一项完整的家庭作业。它们都可以自己正常工作,但是当我将它们放在main函数中时,找到字符串中最小单词的最后一个会产生错误。谁知道为什么?

public static void main(String[] args){
    Scanner sc = new Scanner(System.in);

    //Length of Word
        String word1 = sc.next();
            System.out.println(word1.length());

    //Evens in one string odds in the other
        String word2 = sc.next();
            StringBuilder even = new StringBuilder();
            StringBuilder odd = new StringBuilder();
                for(int i = 0; i < word2.length(); i++){
                    if(i % 2 == 0){
                        even.append(word2.charAt(i));
                    }
                    else{
                        odd.append(word2.charAt(i));
                    }
                }
                System.out.println(even + " " + odd);

    //Diminishing Suffix
        String word3 = sc.next();
            for(int j = 0; j < word3.length(); j++){
                System.out.print(word3.substring(j, word3.length()) + " ");
            }
            System.out.printf("\n");

    //Letter Replacement
        String word4 = sc.next();
        String word5 = sc.next();
        String word6 = sc.next();

        String word7 =  word4.replace(word5, word6);
            System.out.println(word7);

    //How many times x appears in xstring
        String word8 = sc.next();
        String word9 = sc.next();
              int index = word8.indexOf(word9);
              int count = 0;
                while (index != -1) {
                    count++;
                    word8 = word8.substring(index + 1);
                    index = word8.indexOf(word9);
                }
                System.out.println(count);
                System.out.println();

    //Lexicographically smallest word
                String Sentence = sc.nextLine();
                String[] myWords = Sentence.split(" ");
                int shortestLengths, shortestLocation;

                shortestLengths=(myWords[1]).length();

                shortestLocation=1;

                for (int i = 1; i <myWords.length; i++) {

                if ((myWords[i]).length() < shortestLengths) {

                shortestLengths=(myWords[i]).length();

                shortestLocation=i;

                }

                }


                System.out.println(myWords[shortestLocation]);
}

}

Talking about the lexicographically smallest one. Error it gives is:

谈论按字典顺序最小的一个。它给出的错误是:

Exception in thread "main" java.lang.ArrayIndexOutOfBoundsException: 1
at StringFun.main(StringFun.java:77)

2 个解决方案

#1

This line seems to be throwing the exception:

这条线似乎抛出异常:

shortestLengths=(myWords[1]).length();

So myWords must not have a second element, meaning the string Sentence didn't contain a space (" ") to be split on. Since Sentence is coming from sc.nextLine(), the "problem" must be with your input - but the true issue is that you're not handling that case properly.

因此myWords不能有第二个元素,这意味着字符串Sentence不包含要分割的空格(“”)。由于句子来自sc.nextLine(),“问题”必须与您的输入相关 - 但真正的问题是您没有正确处理该案例。

Just glancing at your logic, I don't see why shortestLengths couldn't start out as Integer.MAX_VALUE since you're looping over each word and seeing if its length is less than that.

只是看一下你的逻辑,我不明白为什么shortestLengths不能以Integer.MAX_VALUE开头,因为你循环遍历每个单词并查看它的长度是否小于那个。

#2

If shortestLengths=(myWords[1]).length(); is line 77, I'd say its because your input doesn't have any spaces in it, and you don't check for this. Also on this line, myWords[1] refers to the second word, since 0 is the first index. Your code here ignores the first word entirely.

如果shortestLengths =(myWords [1])。length();是第77行,我会说因为你的输入中没有任何空格,你不检查这个。同样在这一行,myWords [1]引用第二个单词,因为0是第一个索引。这里的代码完全忽略了第一个单词。

#1