java.lang.NumberFormatException: For input string: "11111111111111111111111111111111"

Coderanch | Campbell Ritchie | 6 years ago
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    java.lang.NumberFormatException: For input string: "11111111111111111111111111111111"
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    Stack Overflow | 1 year ago | Codebender
    java.lang.NumberFormatException: For input string: "11111111111111111111111111111111"
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    Why does this code result in a NumberFormatException?

    Stack Overflow | 12 months ago | jcm
    java.lang.NumberFormatException: For input string: "11111111111111111111111111111111"
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Root Cause Analysis

  1. java.lang.NumberFormatException

    For input string: "11111111111111111111111111111111"

    at java.lang.NumberFormatException.forInputString()
  2. Java RT
    Integer.parseInt
    1. java.lang.NumberFormatException.forInputString(NumberFormatException.java:48)
    2. java.lang.Integer.parseInt(Integer.java:461)
    2 frames
  3. Unknown
    IntegerParseDemo.main
    1. IntegerParseDemo.main(IntegerParseDemo.java:5)
    1 frame