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Java Numbers

Edmond Corrola

Joined: May 23, 2008
Posts: 15
I'm reading in positive numbers from a file, I need to do some processing but there's no telling how big these positive numbers will be. So far it looks like up to a max of 10 digits. I've tried declaring different variables to hold them (int, float, long, BigInteger, etc..) but I get a compile time error of "number too large" when I test an integer value of 10 digits on them. What can I use to hold huge numbers?

p.s. I was thinking of just storing them as a string, than processing the parts I need by breaking the string up (because the number represents an address which I need to break into bits). Does this sound like a better solution?
Ernest Friedman-Hill
author and iconoclast

Joined: Jul 08, 2003
Posts: 24199

If it's a compile-time error, that suggests you're writing something like

long x = 123456789012345;

and getting the error. That's because to write a long literal in Java code, you have to mark the actual literal as being a long by appending a letter 'L':

long x = 123456789012345L;

Try it!

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Edmond Corrola

Joined: May 23, 2008
Posts: 15
thanks, in C i always type casted it so i wouldn't have to remember. I completely forgot that I needed to do that. It wasn't making sense that I couldn't use a 10 digit number for a "long".
[ October 24, 2008: Message edited by: Edmond Corrola ]
Rob Spoor

Joined: Oct 27, 2005
Posts: 20276

And if long becomes too small, you can always use java.math.BigInteger. Not as fast but at least it can hold arbitrarily large numbers.

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subject: Java Numbers
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