Convert from one base to another in Java

public class BaseToBaseConv {

static String baseToBase(String num, int base1, int base2) {
    int no = convFrmBaseToDeci(num, base1);
    return convFrmDecToBase(no, base2);
}

static String convFrmDecToBase(int num, int base) {

    String res = "";
    int rem;
    // Convert input number is given base by repeatedly
    // dividing it by base and taking remainder
    while (num > 0) {
        rem = num % base;
        if (base == 16) {
            if (rem == 10)
                res += 'A';
            else if (rem == 11)
                res += 'B';
            else if (rem == 12)
                res += 'C';
            else if (rem == 13)
                res += 'D';
            else if (rem == 14)
                res += 'E';
            else if (rem == 15)
                res += 'F';
            else
                res += rem;
        } else
            res += rem;

        num /= base;
    }
    // Reverse the result
    return new StringBuffer(res).reverse().toString();
}

static int convFrmBaseToDeci(String num, int base) {

    if (base < 2 || (base > 10 && base != 16))
        return -1;

    int val = 0;
    int power = 1;

    for (int i = num.length() - 1; i >= 0; i--) {
        int digit = digitToVal(num.charAt(i));

        if (digit < 0 || digit >= base)
            return -1;

        // Decimal equivalent is str[len-1]*1 +
        // str[len-1]*base + str[len-1]*(base^2) + ...
        val += digit * power;
        power = power * base;
    }

    return val;
}

static int digitToVal(char c) {
    if (c >= '0' && c <= '9')
        return (int) c - '0';
    else
        return (int) c - 'A' + 10;
}

public static void main(String [] args) {
    System.out.println(baseToBase("12345", 10, 2));
    System.out.println(baseToBase("11000000111001", 2, 10));
    System.out.println(baseToBase("ABC11", 16, 2));
    System.out.println(baseToBase("10101011110000010001", 2, 16));
    System.out.println(baseToBase("12322", 8, 16));
}
}

The two-argument versions of Integer.parseInt or Long.parseLong will do this if you can be sure the number in question is within the range of int or long respectively. If you can't guarantee this, use java.math.BigInteger:

BigInteger bi = new BigInteger(number, base1);
return bi.toString(base2);

This can handle arbitrarily-large integers, for example

System.out.println(
  new BigInteger("12345678901234567890123456789", 10).toString(16));
// prints 27e41b3246bec9b16e398115 - too big to represent as a long

I believe this will work:

long x = 10;
int baseToConvertTo = 9;
System.out.println(Long.toString(x, baseToConvertTo));

Output: 11

You could do

return Integer.toString(Integer.parseInt(number, base1), base2);

So with your function signature, in Java:

public String convertFromBaseToBase(String str, int fromBase, int toBase) {
    return Integer.toString(Integer.parseInt(str, fromBase), toBase);
}