Hi,
Noel wrote:
> I fed '26 3 ^p' into 'dc' to see just how big it was - and got
> "17576", a 16-bit word array of which would fit into a PDP-11 64KB
> address space.
I'm a fan of dc(1), but also of units(1) and since both seem underused
here's an alternative method.
$ units -1v '26^3 16 bit' 64KiB
26^3 16 bit = 0.53637695 * 64KiB
--
Cheers, Ralph.