What is the modulo operator for longs in Java?

Question

How do I find the modulo (%) of two long values in Java? My code says 'Integer number too large' followed by the number I'm trying to mod. I tried casting it to a long but it didn't work. Do I have to convert it to a BigInteger and use the remainder method? Thanks.

Solution 1

The % operator does work for longs. It sounds like you may have forgotten to stick L at the end of a numeric literal, as in 123456789L. Can we see your code?

Solution 2

You can only have an integer up to 2 147 483 647. If you want to go bigger than that, say 3 billion, you must specify it to be a long

class Descartes {
    public static void main(String[] args) {
        long orig = Long.MAX_VALUE;
        long mod = orig % 3000000000; // ERROR 3000000000 too big
        long mod = orig % 3000000000L; // no error, specified as a long with the L
    }
}

Keep in mind that you can use capital OR lowercase L, but it's advisable to use capital, since the lowercase looks remarkably similar to the number 1.

Comments

[Descartes]

Okay, the following code is inside a for loop and total is a BigInteger. code total = total.add(BigInteger.valueOf(b(i)%1234567891011) code

[Marvo]

Might you need an L at the end of 1234567891011? Assuming b(i) returns a long as well.

[Daniel Lubarov]

Yeah, just stick an "L" on the end of 1234567891011 to tell the compiler it's a long literal (otherwise the compiler will treat it as an int literal).

[alternative]

His question states that this is already a possibility but he doesn't want to do this.

[elekwent]

You're right. I missed that in his question. I'll leave my answer the way it is though, as sort of an awkward answer to him asking "Do I have to convert it to a BigInteger and use the remainder method?".

[elekwent]

Another way to read my answer above... Yes, you can also try working with the BigInteger class, which has a remainder() method that works similarly to %. You see what I did? I add a yes and two commas, and now it answers his question perfectly.