Java Integer parseInt error

问题

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"

回答1:

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.

回答2:

How about

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

回答3:

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

回答4:

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);

来源：https://stackoverflow.com/questions/6896851/java-integer-parseint-error