NumberFormatException from hex number

Question

In my code I have

int i = Integer.parseInt("f8004896",16);

when I run the program it throws a NumberFormatException

java.lang.NumberFormatException: For input string: "f8004896"

What am I doing wrong? This seems relatively straightforward but its not working.

Yes, does its value fit in an `int`. It's just easier to think in decimal. — Sotirios Delimanolis, Mar 13 '14 at 19:57
@SotiriosDelimanolis - Why is it easier to think in decimal? I find it much easier to remember that the biggest integer is 0x7fffffff than 2147... something. — Dawood ibn Kareem, Mar 13 '14 at 20:04
@DavidWallace Ok, IMO. I just haven't worked with HEX values at all. — Sotirios Delimanolis, Mar 13 '14 at 20:05
The NumberFormatException is kind of misleading when it comes to something like this. The number is formatted correctly it should be like BitwiseOverflowException or something lol — Neil Locketz, Mar 13 '14 at 20:15

score 3 · Accepted Answer · answered Mar 13 '14 at 19:58

3

long i = Long.parseLong("f8004896", 16);
System.out.println(i);
System.out.println(Integer.MAX_VALUE);

Output:

4160768150
2147483647

answered Mar 13 '14 at 19:58

DmitryKanunnikoff

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score 1 · Answer 2 · edited Mar 13 '14 at 20:06

1

Decimal value for f8004896 is 4160768150 and it's more than 2^31-1 (upper limit for int type). So you should use long instead: Long.parseLong()

edited Mar 13 '14 at 20:06

DmitryKanunnikoff

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answered Mar 13 '14 at 20:01

nikis

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You made a mistake in "upper limit for long". I have edited your answer. – DmitryKanunnikoff Mar 13 '14 at 20:09
@DmitryTsechoev yep, it was a typo indeed. Thanks! – nikis Mar 13 '14 at 20:11
Do not mention it. And thank you for your understanding! – DmitryKanunnikoff Mar 13 '14 at 20:13

NumberFormatException from hex number

2 Answers2