Java：使用字符串charAt方法进行循环崩溃 [英] Java: for loop crashing using string charAt method

问题描述

我在这段代码中有一个错误。调试器建议它的原因是这行代码 char chr = getSecretWord.charAt（i）;

这段代码的作用是寻找 userInput 和 secretWord 之间的匹配。我有 for loop 逐个查看secretWord字母的长度，如果有匹配的字母返回true。如果没有，返回false ...但是当假设只返回false时程序崩溃...我想这是这行的东西，但是不知道究竟是什么 getSecretWord.charAt（i） ;

  private boolean isMatchingSecretWord（String userInput）
 {
 String secretWord =; 
 String getSecretWord = getSecretWord（）; 
 for（int i = 0; i< = getSecretWord.length（）; i ++）
 {
 char chr = getSecretWord.charAt（i）; 
 secretWord =+ chr; 
 
 if（secretWord.equals（userInput））
 {
 println（is true）; 
返回true; 
} 
} 
返回false; 
} 
  

作为旁注，我对此代码所做的是正确的，将getSecretWorld（）方法赋给String，这样我就可以使用Strings方法 length（）？

String getSecretWord = getSecretWord（）;

for（int i = 0; i< = getSecretWord.length（）; i ++）

调试代码：

 线程Thread-4中的异常java.lang.StringIndexOutOfBoundsException：字符串索引超出范围：4 
 at java.lang.String.charAt（String.java:686）
 at Hangman.isMatchingSecretWord（Hangman.java:49）
 at Hangman.userInput（Hangman.java:34 ）
在Hangman.run（Hangman.java:20）* 
  

解决方案

  for（int i = 0; i< = getSecretWord.length（）; i ++）
   // ^^^ 
 //见这里
  

n -character字符串（或 n -element数组）的有效索引是 0 到 n-1 包含。

所以，如果您的密码是 xyyzy ，有效索引为0到4。你的原始循环迭代， i 通过 5设置为零，因此问题。

---

但是那里似乎有很多不必要的代码，当你可以逃避简单的事情时。

首先，我会删除一个混乱的来源 - 函数名称听起来像用户输入和秘密字必须完全匹配，而你的评论指示否则：

谢谢，这很有效。但循环的原因是用户输入一个字母，我想看看那封信是否在SecretWord中。 （这是一个刽子手游戏）。

在这种情况下，你只想查看密码中是否存在单个字符。我会更改函数名称以适应，即使这样，也可以使用更少的代码完成：

  private boolean isInSecretWord （String userInput）{
 String secretWord = getSecretWord（）; 
返回secretWord.contains（userInput）; 
} 
  

I have a bug in this block of code. The debugger suggest it´s cause is this line of code char chr = getSecretWord.charAt(i);

What this code does is look for a match between userInput and secretWord. I have the for loop to go through the length of the secretWord letters one by one, and if there is a letter matching return true. If not, return false... but the program crashes when it is suppose to just return false... I guess it is something with this line, but do not know exactly what getSecretWord.charAt(i);

    private boolean isMatchingSecretWord(String userInput)
{
    String secretWord = "";
    String getSecretWord = getSecretWord();
    for (int i = 0; i <= getSecretWord.length();i++)
        {
        char chr = getSecretWord.charAt(i);
        secretWord = ""+chr;

        if (secretWord.equals(userInput))
        {
            println("is true");
            return true;
        }
    }
    return false;
}

As an side note, is what I´ve done with this code correct, assigning the getSecretWorld() Method to a String so I can use the Strings method length()?

String getSecretWord = getSecretWord();

for (int i = 0; i <= getSecretWord.length();i++)

Debug code:

Exception in thread "Thread-4" java.lang.StringIndexOutOfBoundsException: String index out of range: 4    
    at java.lang.String.charAt(String.java:686)    
    at Hangman.isMatchingSecretWord(Hangman.java:49)    
    at Hangman.userInput(Hangman.java:34)    
    at Hangman.run(Hangman.java:20)*

解决方案

for (int i = 0; i <= getSecretWord.length(); i++)

should be:

for (int i = 0; i < getSecretWord.length(); i++)
//               ^^^
//             see here

The valid indexes for an n-character string (or an n-element array) are 0 through n-1 inclusive.

So, if your secret word is xyyzy, the valid indexes are zero through four. Your original loop iterates with i set to zero through five, hence the problem.

---

But there seems to be a lot of unnecessary code in there, when you could get away with something simple.

First, I would remove a source of confusion - the function name sounds like the user input and the secret word have to match completely whereas your comment indicates otherwise:

Thanks, this works. But the reason for the loops is that the user enters one letter, I want to see if that letter is within the SecretWord. (it´s a hangman game).

In that case, you simply want to see if the single character exists in the secret word. I would change the function name to suit and, even then, it can be done with a lot less code:

private boolean isInSecretWord (String userInput) {
    String secretWord = getSecretWord();
    return secretWord.contains(userInput);
}

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