How to take a base 2 logarithm from a boost int256_t?

So I want to take a logarithm from an int256_t.

I found this but modifying it to take sizeof int256_t did not work. It will give incorrect results: https://stackoverflow.com/a/11376759

Is there a log function in boost which supports multiprecision?

Solution

Without further constraints I'd write:

int256_t x("12345678901234567890");

std::cout << "log2(" << x << ") = " << log2(cpp_bin_float_100(x)) << "\n";

Prints

log2(12345678901234567890) = 63.4206

Simpler

You don't need all that precision if you're going to round the result anyways:

std::cout << "log2(" << x << ") = " << log2(x.convert_to<double>()) << "\n";

A very crude form could look like:

template <typename T>
static inline int adhoc_log2(T i) {
    int n = 0;
    while(i /= 2) ++n;
    return n;
}

Prints

adhoc(12345678901234567890) = 63

Lastly, indeed you could drop back down to bit-level and apply the bit-fiddling hacks in the way @SamVarshavchik suggests