Long.parseLong的NumberFormatException

Question

when I try to run this Code (yes, I know it will run horribly long and is not very pretty):

DecimalFormat formatA = new DecimalFormat("000");
        DecimalFormat formatB = new DecimalFormat("0000000000000000000000000000000000000000");

        for(int c=0; c<=7; c++){
            String binC = formatA.format(Integer.parseInt(Integer.toBinaryString(c)));
            for(int d=0; d<LFSRa.length; d++){
                LFSRa[d]=Character.getNumericValue((binC.charAt(d)));                   
            }
            for(long e=0; e<=1099511627775L; e++){
                String binE = formatB.format(Long.parseLong((Long.toBinaryString(e))));
                for(int f=0; f<LFSRb.length; f++){
                    LFSRb[f]=Character.getNumericValue((binE.charAt(f)));
                }


            }
        }

I get the following exception

Exception in thread "main" java.lang.NumberFormatException: For input string: "10000000000000000000"
at java.lang.NumberFormatException.forInputString(Unknown Source)
at java.lang.Long.parseLong(Unknown Source)
at java.lang.Long.parseLong(Unknown Source)
at A41.main(A41.java:76)

and I don't get why. What should I do? Could somebody explain the problem to me?

Answer 1

if you see this:

http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

The range of long ie max is 9223372036854775807, 19 digits and once it reaches 10000000000000000000 (20 digits), it will try to convert this string to long which is out of range and hence you get an exception.