Calculating powers of integers

When it's power of 2. Take in mind, that you can use simple and fast shift expression 1 << exponent

example:

2² = 1 << 2 = (int) Math.pow(2, 2)
2¹⁰ = 1 << 10 = (int) Math.pow(2, 10)

For larger exponents (over 31) use long instead

2³² = 1L << 32 = (long) Math.pow(2, 32)

btw. in Kotlin you have shl instead of << so

(java) 1L << 32 = 1L shl 32 (kotlin)

Integers are only 32 bits. This means that its max value is 2^31 -1. As you see, for very small numbers, you quickly have a result which can't be represented by an integer anymore. That's why Math.pow uses double.

If you want arbitrary integer precision, use BigInteger.pow. But it's of course less efficient.

Best the algorithm is based on the recursive power definition of a^b.

long pow (long a, int b)
{
    if ( b == 0)        return 1;
    if ( b == 1)        return a;
    if (isEven( b ))    return     pow ( a * a, b/2); //even a=(a^2)^b/2
    else                return a * pow ( a * a, b/2); //odd  a=a*(a^2)^b/2

}

Running time of the operation is O(logb). Reference:More information