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Integer.parseString throws java.lang.NumberFormatException

Posted on 2004-09-14
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1,802 Views
Last Modified: 2008-03-06
Want to know why I get the above error in the following program for string s2. The string s1 and s2 are hex strings.

public class sample1{
public static void main(String args[])
     {
       String s1 = "6666B5C2";
       String s2 = "85EB3EC2";

       int val1 = Integer.parseInt(s1,16);
       System.out.println("The value for s1 is : " + val1);

       int val2 = Integer.parseInt(s2,16);
       System.out.println("The value for s2 is : " + val2);
     }
}

Thanks,
PK
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Question by:pkhanna01
1 Comment
 
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Accepted Solution

by:
KeithWatson earned 125 total points
ID: 12060744
The second number in decimal is 2,246,786,754, which is just too big to fit into a 32 bit signed int. It will need to be a maximum of 2,147,483,647 to fit.
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