Are these Java long integer or BigInteger format numbers? IP2Location csv database

Question

I have never used long integer or BigInteger format numbers in my Java code, and this aspect of the IP2Location databases does not appear to be documented. I need to know which it is, so that I can write some simple code to compare numbers for greater that or equal. Here are a couple of lines of data from the file...

"281470698522624","281470698524671","CN","China","Guangdong","Guangzhou"
"281470698524672","281470698528767","JP","Japan","Tokyo","Tokyo"

EDIT: The two numbers at the beginning of each line represent a range of IP addresses that are located in the city identified by the last 4 entries on the line. I convert my IP address to a decimal notation following a known algorithm, then search thru the file sequentially until I find a second number that is greater than or equal to my IP. Then I have the location data :) QED

Hope you can help, Mick :)

Answer 1

It is documented:

ip_from INT(10)  / Decimal(39,0)    First IP address in netblock.
ip_to   INT(10)  / Decimal(39,0)    Last IP address in netblock.

Given that your samples consist of more than 10 digits, you have the IPv6 version of DB4. Which would have been a good thing to mention in your question. These numbers can grow very large - larger than long can support (which can go as far as Long.MAX_VALUE, which is 2^63-1, far less than 39 nines, which is as far as these numbers can theoretically go.

You could represent them with BigInteger, or 2 longs (IPv6 addresses are 128 bits long, and longs are 64 bit, so 2 longs can do it - given that this is basic IPv6 info, I don't think it's useful or necessary for IP2Location to mention this. In fact, calling it 'decimal(39,0)' is a bit of a daft way to state 'IPv6 address as a single number stated in decimal'. At any rate, they represent InetAddresses, so you might want to use the class that actually describes what it is, namely, InetAddress.

Answer 2

I found methods for everything that I need in this...

IPAddress Java library
https://seancfoley.github.io/IPAddress/

The manager of this library has posted this link in a few of the threads on this topic.

Answer 3

You asked:

Are these Java long integer or BigInteger format numbers?

They are long values.

In Java, you can see the min and max by using setting MIN_VALUE and MAX_VALUE for each of Byte, Short, Integer, Long, Float, and Double. You can see how many bits wide they are by using SIZE.

Examples:

System.out.println(Long.MAX_VALUE);
System.out.println(Long.SIZE);

prints

9223372036854775807
64

Check out Double and Float for more constants such as MIN_EXPONENT and PRECISION.

For requirements exceeding those types, check out BigInteger and BigDecimal

Also, if you want to find the maximum number of digits, in a long (similar for int, short, or byte, you can do the following:

double digitsInLong = (Long.SIZE*Math.log(2))/Math.log(10);
System.out.println(digitsInLong);

prints

19.265919722494793

This says there can be between 19 and 20 digits. All longs can have 19 digits but some longs with 20 will overflow the 63 bit size limit for positive numbers.

Answer 4

You can use the Long#parseLong method, to convert a String value to a long primitive data-type.

String[] a = { "281470698522624","281470698524671","CN","China","Guangdong","Guangzhou" };
String[] b = { "281470698524672","281470698528767","JP","Japan","Tokyo","Tokyo" };
long a0 = Long.parseLong(a[0]);
long a1 = Long.parseLong(a[1]);
long b0 = Long.parseLong(b[0]);
long b1 = Long.parseLong(b[1]);
System.out.println("a0 = " + a0);
System.out.println("a1 = " + a1);
System.out.println("b0 = " + b0);
System.out.println("b1 = " + b1);

Output

a0 = 281470698522624
a1 = 281470698524671
b0 = 281470698524672
b1 = 281470698528767

If you'd like to use the BigInteger class, you can simply supply the String value as the parameter of the constructor method.

BigInteger a0 = new BigInteger(a[0]);
BigInteger a1 = new BigInteger(a[1]);
BigInteger b0 = new BigInteger(b[0]);
BigInteger b1 = new BigInteger(b[1]);