- not only multiple of 8. We can do it by using uint8 typed arrays, and
it does not hurt performance:

In [1]: from wendelin.lib.mem import bzero, memset, memcpy
In [2]: A = bytearray(2*1024*1024)
In [3]: B = bytearray(2*1024*1024)

        memcpy(B, A)    bzero(A)        memset(A, 0xff)

old:    718 µs          227 µs / 1116   228 µs / 1055 (*)
new:    718 µs          176 µs / 1080   175 µs / 1048

    (*) the second number comes from e.g.

        In [8]: timeit bzero(A)
        The slowest run took 4.63 times longer than the fastest.
        This could mean that an intermediate result is being cached
        10000 loops, best of 3: 228 µs per loop

        so the second number is more realistic and says performance
        stays aproximately the same and only slightly improves.