java.lang.NumberFormatException: For input string: "a6acda419199c58a"

Coderanch | Richard Johnson | 1 decade ago
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  1. 0

    is this a bug with Long.parseLong() code?

    Coderanch | 1 decade ago | Richard Johnson
    java.lang.NumberFormatException: For input string: "a6acda419199c58a"
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    Checking if string contains only allowed characters

    Stack Overflow | 3 years ago | Maciej Socha
    java.lang.NumberFormatException: Invalid long: "112a"
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    java.lang.NumberFormatException: For input string: “10.00”

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Root Cause Analysis

  1. java.lang.NumberFormatException

    For input string: "a6acda419199c58a"

    at java.lang.NumberFormatException.forInputString()
  2. Java RT
    Long.parseLong
    1. java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
    2. java.lang.Long.parseLong(Long.java:406)
    2 frames