Calculate exponents of 16 using only bitwise operators

Question

If I was using a loop that looked like this:

String string = "DF3";
for (int i = string.length() - 1; i >= 0; --i) {
        int c = string.charAt(i);

        convertedHex += (i << 4);

}

How would I calculate Math.pow(16, i) without using Math.pow() , another loop, or multiplication? Specifically, how can I do it with only bitwise operators?

Answer 1

You can convert i -th power of 16 by observing that 16=2 ⁴ , hence 16 ⁱ =2 ^4*i , and that 2 ⁱ = 1<<i . Therefore, 16 ⁱ = 1<<i*4 , which can be rewritten as 1<<(i<<2) to avoid multiplication.

However, you do not need to compute the powers of 16 directly. You can construct the number by repeated multiplication by 16 in the loop, ie

convertedHex = (convertedHex << 4) + nextHexDigit;

Note: int c = string.charAt(i) gives you the UNICODE code point for the character, not the value of the corresponding digit. Use this code instead:

int nextHexDigit = Character.digit(string.charAt(i), 16);

Answer 2

Given that 1 << i is same as Math.pow(2, i) , you can simply use 1 << (i * 4) . However, note that this is assuming that overflow does not occur for the datatype you are using.

In case of int , a maximum value of 7 is allowed for i, and in case of long , a maximum value of 15.