hiiii frns...


integer can hold only upto 8 bits...where as a string can hold more than 8 bits...so to compare them...is der any function???

like my code is like this

for(int i=0;i<= stringgg("12345678901234567890"); i++)

1. Java integers are 32 bits.
2. Java Strings are 16 bits per character

3. To compare them or use the String in arithmetic expressions, you first need to convert the string to integer with something like int i = new Integer("12345");

An int is 4 bytes which is 32 bits (31 of which is value and the 32nd is a pos/neg toggle). A long is 8 bytes (63 bits data and 1 pos/neg toggle) so use long and parse that String to a long using Long.parseLong(string). If a long is also not large enough use the BigDecimal class. See the API docs.

An integer in Java is a 32-bit number. If you need anything larger than that, I suggest you look at the Math.BigInteger class. Documentation for a BigInteger can be found here.

Looks like James, masijade and I all posted at the same time, and I forgot about the long type Use that as masijade said.

and since you're working with String-objects, make sure they're valid numeric values before you try to compare them

Ach, yeah, and as darkagn say, BigInteger, not BigDecimal (unless you're going to have decimals, in which case you'd have been using float and double instead of int and long, anyway).

Yes, long is necessary if your values can exceed 2^31 (although I suspect your example loop was just an example; you don't really want to execute a loop that many times!).
If your String isn't valid for converting to integer (or long) you'll get a NumberFormatException thrown.

thanks for the fast responses!!!!

but is their any other methods other than big integers...

What do u need exactly?