How to handle arbitrarily large integers

I'm working on a programming language, and today I got the point where I could compile the factorial function(recursive), however due to the maximum size of an integer the largest I can get is factorial(12). What are some techniques for handling integers of an arbitrary maximum size. The language currently works by translating code to C++.

If you need larger than 32-bits you could consider using 64-bit integers (long long), or use or write an arbitrary precision math library, e.g. GNU MP.

If you want to roll your own arbitrary precision library, see Knuth's Seminumerical Algorithms, volume 2 of his magnum opus.

If you're building this into a language (for learning purposes I'd guess), I think I would probably write a little BCD library. Just store your BCD numbers inside byte arrays.

In fact, with today's gigantic storage abilities, you might just use a byte array where each byte just holds a digit (0-9). You then write your own routine to add, subtract multiply and divide your byte arrays.

(Divide is the hard one, but I bet you can find some code out there somewhere.)

I can give you some Java-like psuedocode but can't really do C++ from scratch at this point:

class BigAssNumber {
    private byte[] value;

    // This constructor can handle numbers where overflows have occurred.
    public BigAssNumber(byte[] value) {
        this.value=normalize(value);
    }

    // Adds two numbers and returns the sum.  Originals not changed.
    public BigAssNumber add(BigAssNumber other) {
        // This needs to be a byte by byte copy in newly allocated space, not pointer copy!
        byte[] dest = value.length > other.length ? value : other.value;         

        // Just add each pair of numbers, like in a pencil and paper addition problem.
        for(int i=0; i<min(value.length, other.value.length); i++)
            dest[i]=value[i]+other.value[i];

        // constructor will fix overflows.
        return new BigAssNumber(dest);
    }

    // Fix things that might have overflowed  0,17,22 will turn into 1,9,2        
    private byte[] normalize(byte [] value) {
        if (most significant digit of value is not zero)
            extend the byte array by a few zero bytes in the front (MSB) position.

        // Simple cheap adjust.  Could lose inner loop easily if It mattered.
        for(int i=0;i<value.length;i++)
            while(value[i] > 9) {
                value[i] -=10;
                value[i+1] +=1;
            }
        }
    }
}

I use the fact that we have a lot of extra room in a byte to help deal with addition overflows in a generic way. Can work for subtraction too, and help on some multiplies.