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?
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1It may look like binary, that is going to be interpreted as decimal (and too high).pamphlet– pamphlet2014-02-07 20:42:10 +00:00Commented Feb 7, 2014 at 20:42
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2 Answers
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)
3 Comments
Harald K
As OP says "binary digits", the second version sounds like the right choice. :-)
Maximii77
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.
Zinglish
In that case parse it to Long, just like the first example in the answer.
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.
