java type_error ai_generated true

java.util.InputMismatchException

ID: java/inputmismatchexception

Also available as: JSON · Markdown · 中文
88%Fix Rate
86%Confidence
1Evidence
2023-06-18First Seen

Version Compatibility

VersionStatusIntroducedDeprecatedNotes
Java 8 active — — —
Java 11 active — — —
Java 17 active — — —
Java 21 active — — —

Root Cause

The input token does not match the expected type when using a Scanner, often due to incorrect locale or unexpected data format.

generic

中文

使用 Scanner 时输入令牌与期望类型不匹配，通常由于区域设置错误或数据格式意外。

Official Documentation

https://docs.oracle.com/en/java/javase/17/docs/api/java.base/java/util/InputMismatchException.html

Workarounds

  1. 90% success Use hasNextInt() before nextInt(): 'if (scanner.hasNextInt()) { int value = scanner.nextInt(); } else { scanner.next(); /* consume invalid token */ }' 
    Use hasNextInt() before nextInt(): 'if (scanner.hasNextInt()) { int value = scanner.nextInt(); } else { scanner.next(); /* consume invalid token */ }'
  2. 85% success Set the locale to match the input format: 'scanner.useLocale(Locale.US);' for decimal numbers with dot separator. 
    Set the locale to match the input format: 'scanner.useLocale(Locale.US);' for decimal numbers with dot separator.

中文步骤

  1. Use hasNextInt() before nextInt(): 'if (scanner.hasNextInt()) { int value = scanner.nextInt(); } else { scanner.next(); /* consume invalid token */ }'
  2. Set the locale to match the input format: 'scanner.useLocale(Locale.US);' for decimal numbers with dot separator.

Dead Ends

Common approaches that don't work:

  1. 60% fail

    Setting the locale to US (Locale.US) may not fix the issue if the input contains non-numeric characters.

  2. 70% fail

    Using next() instead of nextInt() and then manually parsing can still throw NumberFormatException if the string is malformed.