Java Long parselong throwing Number Format exception

Question

I have a simple code first I use this

Long.parseLong(4250120140405520712)

And it works but when I do this

Long.parseLong(42501201404055207123)

It fails. Adding an extra digit makes it throw a Number Format Exception. Can someone please explain

score 4 · Answer 1 · answered Sep 18 '14 at 21:14

4

Assuming you're parsing Strings into longs:

The first one works because the number 4250120140405520712 (19 digits) is less than the maximum possible long value, Long.MAX_VALUE, 9223372036854775807L.

The second one fails because it's 20 digits long, bigger than 9223372036854775807L.

answered Sep 18 '14 at 21:14

rgettman

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and we can use BigInteger to represent the second number – Paraneetharan Saravanaperumal Jun 08 '15 at 05:48

Ḟḹáḿíṅḡ Ⱬỏḿƀíé · Answer 2 · 2014-09-18T21:45:21.410

1

long variables can only hold a maximum of 9223372036854775807 see here for more information.
The reason the second one doesn't work is that you have exceeded the limit so the number format exception is shown.

edited Sep 18 '14 at 21:45

answered Sep 18 '14 at 21:16

Ḟḹáḿíṅḡ Ⱬỏḿƀíé

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Java Long parselong throwing Number Format exception

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