java.lang.NumberFormatException: For input string: "11000000000000000000000000000011"

Coderanch | nirjari patel | 6 years ago
  1. 0

    decimal to binary

    Coderanch | 6 years ago | nirjari patel
    java.lang.NumberFormatException: For input string: "11000000000000000000000000000011"
  2. 0

    Number Fortmat Exception for input string: "FF" (in java.lang.numberfo....)

    Oracle Community | 9 years ago | 843859
    java.lang.NumberFormatException: For input string: "FF"
  3. 0

    Java: formatting numbers error?

    Stack Overflow | 2 years ago | dzenesiz
    java.lang.NumberFormatException: For input string: "C"
  4. Speed up your debug routine!

    Automated exception search integrated into your IDE

  5. 0

    convert hexadecimal number to binary

    Stack Overflow | 2 years ago | Sachithra Sewwandi
    java.lang.NumberFormatException: For input string: "e24dd004"
  6. 0

    How do I open hex file in java?

    Stack Overflow | 4 years ago | zarcel
    java.lang.NumberFormatException: For input string: "tîxl¸?

  1. Handemelindo 6 times, last 5 days ago
  2. bmacedo 2 times, last 7 days ago
  3. MoYapro 1 times, last 1 week ago
  4. filpgame 4 times, last 3 weeks ago
  5. silex 6 times, last 1 month ago
32 more registered users
63 unregistered visitors
Not finding the right solution?
Take a tour to get the most out of Samebug.

Tired of useless tips?

Automated exception search integrated into your IDE

Root Cause Analysis

  1. java.lang.NumberFormatException

    For input string: "11000000000000000000000000000011"

    at java.lang.NumberFormatException.forInputString()
  2. Java RT
    Integer.parseInt
    1. java.lang.NumberFormatException.forInputString(Unknown Source)
    2. java.lang.Integer.parseInt(Unknown Source)
    2 frames
  3. package1
    JigBitOp.main
    1. package1.JigBitOp.main(JigBitOp.java:25)
    1 frame