4 Mar
2023
4 Mar
'23
3:22 p.m.
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.