java中的字符串索引越界错误（charAt）
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string index out of bounds error in java (charAt)
提问by Kyle
Quick question. I have this code in a program:
快速提问。我在一个程序中有这个代码：
input = JOptionPane.showInputDialog("Enter any word below")
int i = 0;
for (int j = 0; j <= input.length(); j++)
{
System.out.print(input.charAt(i));
System.out.print(" "); //don't ask about this.
i++;
}
- Input being user input
ibeing integer with value of 0, as seen
- 输入是用户输入
i为整数，值为 0，如所见
Running the code produces this error:
运行代码会产生这个错误：
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String index out of range: 6
at java.lang.String.charAt(Unknown Source)
at program.main(program.java:15)
线程“main”中的异常 java.lang.StringIndexOutOfBoundsException: String index out of range: 6
at java.lang.String.charAt(Unknown Source)
at program.main(program.java:15)
If I change the charAtintto 0 instead of i, it does not produce the error...
what can be done? What is the problem?
如果我将 更改charAtint为 0 而不是i，它不会产生错误......
可以做什么？问题是什么？
采纳答案by Mena
Replace:
代替：
j <= input.length()
... with ...
... 和 ...
j < input.length()
Java Stringcharacter indexing is 0-based, so your loop termination condition should be at input's length - 1.
JavaString字符索引是基于 0 的，因此您的循环终止条件应该是 atinput的长度 - 1。
Currently, when your loop reaches the penultimate iteration before termination, it references inputcharacter at an index equal to input's length, which throws the StringIndexOutOfBoundsException(a RuntimeException).
目前，当您的循环在终止前到达倒数第二次迭代时，它会input在等于input的长度的索引处引用字符，这会抛出StringIndexOutOfBoundsException(a RuntimeException)。
回答by Guillermo Merino
You were accessing the array from [0-length], you should do it from [0-(length-1)]
你从 [0-length] 访问数组，你应该从 [0-(length-1)]
int i = 0;
for (int j = 0; j < input.length(); j++)
{
System.out.print(input.charAt(i));
System.out.print(" "); //don't ask about this.
i++;
}
回答by Benjamin
Try the following:
请尝试以下操作：
j< input.length()
and then:
进而：
int i = 0;
for (int j = 0; j < input.length(); j++)
{
System.out.print(input.charAt(i));
System.out.print(" "); //don't ask about this.
i++;
}
回答by Hüseyin BABAL
Use this;
用这个;
for (int j = 0; j < input.length(); j++)
{
System.out.print(input.charAt(j));
System.out.print(" "); //don't ask about this.
}
回答by Bathsheba
String indexing in Java (like any other array-like structure) is zero-based. This means that input.charAt(0)is the leftmost character. The last character is then at input.charAt(input.length() - 1).
Java 中的字符串索引（与任何其他类似数组的结构一样）是从零开始的。这意味着这input.charAt(0)是最左边的字符。最后一个字符位于input.charAt(input.length() - 1)。
So you are referencing one too many elements in your forloop. Replace <=with <to fix. The alternative (<= input.length() - 1) could bite you hard if you ever port your code to C++ (which has unsigned types).
因此，您在for循环中引用了太多元素。替换<=与<要修复。<= input.length() - 1如果您将代码移植到 C++（具有无符号类型），则替代 ( ) 可能会让您大吃一惊。
By the way, the Java runtime emits extremely helpful exceptions and error messages. Do learn to read and understand them.
顺便说一下，Java 运行时会发出非常有用的异常和错误消息。一定要学会阅读和理解它们。
回答by nikis
for (int j = 0; j < input.length(); j++)
{
System.out.print(input.charAt(j));
System.out.print(" "); //don't ask about this.
}
回答by Bhujang Bhagas
Replace for loop condition j <= input.length()with j < input.length(), as string in Java follows zero based indexing.
e.g. indexing for the String "india"would start from 0 to 4.
更换for循环的条件j <= input.length()有j < input.length()，如字符串在Java中如下从零开始的索引。例如，字符串的索引"india"将从 0 开始到 4。
