I ran out of labels and registers so here is a way to calculate the number of days per month using only one flag. Note that it always returns 28 days for February, so it's 98% accurate ;)
Code:
CF 0
2
x><y
TEST x = y
SF 0
.
8
8
/
INT
2
GSB "modulo" ; you need a modulo function somewhere
3
0
+
F? 0
-
ABS
RTN
As you can see you need a modulo function. If you don't already have one in your calculator you can use this one:
Code:
LBL "modulo"
X<>Y
STO .7
X<>Y
/
LastX
X<>Y
INT
x
RCL .7
X<>Y
–
RTN
(04-07-2018 12:50 PM)michaelzinn Wrote: [ -> ]I ran out of labels and registers so here is a way to calculate the number of days per month using only one flag. Note that it always returns 28 days for February, so it's 98% accurate 
That's a nice idea: dividing by 0,88 adds 1 to the integer part of a month ≥ 8.
But there is no need for a special mod routine: since only x mod 2 is calculated you can do it with a simple 2 / FRAC 2 *.
Here is another method that uses only integer arithmetics so that it can be used on systems without floating point support:
days = ((m + m shr 3) and 1) + 30
The shift right operation divides by 8 and leaves the integer part (0 or 1), and the AND operator returns 0 for even and 1 for odd numbers.
Maybe someone can implement this on a 16C.
Yes, the February case has to be handled separately. ;-)
A few more thoughts on this – with more or less practical benefit – can be found
here. Including the February case and leap year detection.
Dieter