How does UDOT handle space?
tiborvass opened this issue · 2 comments
tiborvass commented
In your blogpost you mention:
The space is printed by pushing a fake "digit" that will get converted into a space.
but if I read correctly https://github.com/NieDzejkob/miniforth/blob/2842b9a303c07d8d8ace8dc890c531d55866348e/boot.s#L326-L345
You go:
- from
" " - "0" = 32 - 48 = -16
https://github.com/NieDzejkob/miniforth/blob/2842b9a303c07d8d8ace8dc890c531d55866348e/boot.s#L328 - to
-16 + "0" = 32
https://github.com/NieDzejkob/miniforth/blob/master/boot.s#L337 - to
32 + "A" - "0" - 10 = 32 + 65 - 48 - 10 = 39 ≠ 32
https://github.com/NieDzejkob/miniforth/blob/2842b9a303c07d8d8ace8dc890c531d55866348e/boot.s#L340
Am I missing something or should "A" - "0" - 10
be "A" - "0" - 17
?
meithecatte commented
The part that that adds "A" - "0" - 10 won't run because 32 is less than
"9" = 57.
tiborvass commented
🤦 I need to sleep !!! Sorry!