Java Integer parseInt error

Question

I have following problem:

I want to convert some Binary Strings to an integer:

eargb = Integer.parseInt(al + re + gre + blu, 2);

but I get following exception. Why?

java.lang.NumberFormatException: For input string: "11111111111000101000100111111010"

Mark Byers · Accepted Answer · 2011-08-01 10:06:33Z

8

Your number (4,293,036,538) is too large to fit in a signed int (which has a range of -2,147,483,648 to 2,147,483,647).

Try using a long instead. This has a larger range.

answered Aug 1, 2011 at 10:06

Mark Byers

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Comments

adarshr · Accepted Answer · 2011-08-01 10:08:09Z

3

How about

long eargb = Long.parseLong(al + re + gre + blu, 2);

answered Aug 1, 2011 at 10:08

adarshr

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nidhin · Accepted Answer · 2011-08-01 10:07:03Z

1

Your binary number exceeded Integer size. Thats why your getting this exception

answered Aug 1, 2011 at 10:07

nidhin

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Sumit Singh · Accepted Answer · 2012-10-26 05:12:46Z

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It has been 7 months but the target answer has not been described. Also this question is leading in search engines. The above mentioned subjects are true. You should use as follow:

(int)Long.parseLong("11111111111000101000100111111010",2)

eargb =(int)Long.parseLong( al + re + gre + blu, 2);

edited Oct 26, 2012 at 5:12

Sumit Singh

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answered Mar 24, 2012 at 21:50

Mesut Can Gürle

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