Java String to Integer Conversion throws NumberFormatException

Question

I want to cast a string (binary digits) to an Integer like this:

Integer.parseInt("011000010110")

I always get an NumberFormatException. Is the number of digits too high?

It may look like binary, that is going to be interpreted as decimal (and too high). — pamphlet, Feb 07 '14 at 20:42

score 8 · Answer 1 · answered Feb 07 '14 at 20:42

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Yes, the string "011000010110" is about 11 billion, which is higher than the maximum representable int, Integer.MAX_VALUE, 2,147,483,647. Try

Long.parseLong("011000010110")

Or, if you meant it as binary, pass a radix of 2 to parseInt:

Integer.parseInt("011000010110", 2)

answered Feb 07 '14 at 20:42

rgettman

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As OP says "binary digits", the second version sounds like the right choice. :-) – Harald K Feb 07 '14 at 20:48
I´m confused. if I print the parseInt("binary",2) return the decimal value is printed. I wanted the binary digits just as an integer. Do you understand me? Just the "digits" as halraldK said, not the value. – Maximii77 Feb 07 '14 at 21:40
In that case parse it to Long, just like the first example in the answer. – Zinglish Feb 08 '14 at 20:58

score 0 · Answer 2 · edited May 23 '17 at 12:05

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All of Java's number classes are base 10. However, I found these two options here:

working with binary numbers in java

The BitSet class, or a way to declare an int as a binary number (Java 7+). The latter may not work for you depending on how you're obtaining these numbers.

edited May 23 '17 at 12:05

Community

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answered Feb 08 '14 at 20:56

user1854994

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Java String to Integer Conversion throws NumberFormatException

2 Answers2