【发布时间】：2019-05-25 22:11:05
【问题描述】：
我正在尝试编写一个接受用户输入并将其转换为莫尔斯电码的应用程序
功能代码：
public static String[] stringToMorseSequence(String stringOfLetters){
int lenStringOfLetters = stringOfLetters.length();
String output[] = new String[1];
String compInterpreted = "";
for(int i = 0; i < lenStringOfLetters; i++) {
String focus = String.valueOf(stringOfLetters.charAt(i));
String nextCharacter = "";
if (i != lenStringOfLetters) {
nextCharacter = String.valueOf(stringOfLetters.charAt(i + 1));
} else {
nextCharacter = "END";
}
if (focus == " ") {
compInterpreted = compInterpreted + "#";
} else {
String morseSequence = processCharacter(focus);
if (morseSequence == "") {
} else if (i == lenStringOfLetters) {
compInterpreted = compInterpreted + "!";
} else if (nextCharacter != "END") {
if (nextCharacter != " "){
compInterpreted = compInterpreted + morseSequence + "@";
} else {
compInterpreted = compInterpreted + morseSequence;
}
}
}
}
Log.i("MORSE","Computer interpretable morse sequence = " + compInterpreted);
output[0] = compInterpreted;
String prettySequence = "";
for(int i = 0; i-1 < output[0].length(); i++) {
switch(String.valueOf(output[0].charAt(i))){
case("#"):
prettySequence = prettySequence + " ";
case("@"):
prettySequence = prettySequence + " ";
default:
prettySequence = prettySequence + output[0].charAt(i);
}
}
Log.i("MORSE","Human interpretable morse sequence = " + prettySequence);
output[1] = prettySequence;
return output;
}
我似乎收到了这个异常
E/AndroidRuntime: FATAL EXCEPTION: main
Process: me.merhlim.jessica.morsecode, PID: 5687
java.lang.StringIndexOutOfBoundsException: length=10; index=10
at java.lang.String.charAt(Native Method)
at me.merhlim.jessica.morsecode.MorseProcessing.stringToMorseSequence(MorseProcessing.java:16)
at me.merhlim.jessica.morsecode.texttomorse$7.onClick(texttomorse.java:112)
at android.view.View.performClick(View.java:6294)
at android.view.View$PerformClick.run(View.java:24770)
at android.os.Handler.handleCallback(Handler.java:790)
at android.os.Handler.dispatchMessage(Handler.java:99)
at android.os.Looper.loop(Looper.java:164)
at android.app.ActivityThread.main(ActivityThread.java:6494)
at java.lang.reflect.Method.invoke(Native Method)
at com.android.internal.os.RuntimeInit$MethodAndArgsCaller.run(RuntimeInit.java:440)
at com.android.internal.os.ZygoteInit.main(ZygoteInit.java:807)
at de.robv.android.xposed.XposedBridge.main(XposedBridge.java:108)
我知道这是一个逻辑错误，但我不确定这个错误在哪里，或者如何解决它。如错误消息所述，尝试解析字符串中的特定字符时出现索引错误，但我不确定要触发此错误的逻辑不正确
我是 java 新手，而且是深夜，所以也许我只是没有完全理解我正在编写的逻辑中的问题
如果这些信息不足以回答我的问题，我可以提供更多信息
感谢您抽出宝贵时间阅读本文，如果您能提供帮助，我们将不胜感激 -杰西卡
【问题讨论】：
-
提示：
i != lenStringOfLetters永远为真。 -
另外，否则 if (i == lenStringOfLetters) 永远不会为真。
-
线索在异常文本中。字符串的长度为
length=10，您要求的索引为index=10。由于索引是从 0 开始的，因此与长度相同或大于长度的索引会导致此异常。