Presumably, iterating over x>>1 until x == 0 should do the trick? --Tim
On Tuesday, October 06, 2015 08:21:35 AM Scott Jones wrote: > Sorry! I should have been more specific. > What I want is: > 0 -> 0 > 1 -> 1 > 0x8 -> 4 > 0x1f -> 5 > 13 -> 5 > > i.e. number of bits needed to represent the number. > I want to pack 2 or 3 values into a unsigned int, (maybe a UInt16, up to a > UInt128), that I can then use for sorting purposes efficiently. > (much more cache efficient, eliminates a bunch of pointer references, etc.) > > Any idea how? (normally, I'd just use the assembly instructions available > that do this, but I want to do this in pure Julia [it would be nice if the > Julia code > could actually be smart enough to generate the correct native code ;-) ) > > Thanks! > > On Tuesday, October 6, 2015 at 11:03:03 AM UTC-4, Steven G. Johnson wrote: > > On Tuesday, October 6, 2015 at 10:59:04 AM UTC-4, Scott Jones wrote: > >> I couldn't find anything yet - is there a recommended / fastest way to > >> get the number of bits in a number (I really only need it for unsigned > >> values). > >> Thanks > > > > sizeof(number)*8 if you want all the bits (though you'd need to define a > > separate method for BigInt), or count_ones(number) if you want the 1 bits.
