Compile time computing of number of bits needed to encode n different states

Try this:

constexpr unsigned numberOfBits(unsigned x)
{
    return x < 2 ? x : 1+numberOfBits(x >> 1);
}

Simpler expression, correct result.

EDIT: "correct result" as in "the proposed algorithm doesn't even come close"; of course, I'm computing the "number of bits to represent value x"; subtract 1 from the argument if you want to know how many bits to count from 0 to x-1. To represent 1024 you need 11 bits, to count from 0 to 1023 (1024 states) you need 10.

EDIT 2: renamed the function to avoid confusion.

The minimum number of bits required to store n different states is ceil(log2(n)).

constexpr unsigned floorlog2(unsigned x)
{
    return x == 1 ? 0 : 1+floorlog2(x >> 1);
}

constexpr unsigned ceillog2(unsigned x)
{
    return x == 1 ? 0 : floorlog2(x - 1) + 1;
}

Note that ceillog2(1) == 0. This perfectly fine, because if you want to serialize an object, and you know that one of its data members can only take on the value 42, you don't need to store anything for this member. Just assign 42 on deserialization.