Leap days (or not) in February for PBDOS

Leap days (or not) in February for PBDOS - Printable Version

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Leap days (or not) in February for PBDOS - Dale Yarker - 08.03.2025

'DECLARE not needed if FUNCTION source is before the CALL
'here the CALL is in PBMAIN
'I call it bottom up because PBMAIN is last and flow goes up
'
'Pretty sure PUSHes and POPs not needed in FUNCTION
'They will be needed to use "inside" code in-line with BASIC
'
'Used PBCC v6 to test. I no longer have PBDOS available.
function Tage_im_Februar(byval iJahr as integer) as integer
local FebruarTage as integer
! push ax
! mov FebruarTage, 28& 'not a leap year, pre-set to 28
! mov ax, 1582%
! cmp ax, iJahr
! jge Julian
! mov ax, 3%
! and ax, iJahr 'conditionally equivalent MOD 4
! jz MOD100 '0 is possible leap year
! jmp Done
MOD100:
! push bx 'for divisor
! push dx 'for high part of dividend, remainder (MOD)
! xor dx, dx
! mov ax, iJahr
! mov bx, 100%
! div bx
! cmp dx, 0% 'does MOD 100 = 0 ?
! pop dx
! pop bx
! jz MOD400 '0 is possibly not a leap year
! jmp Is29Days 'non 0 is a leap year
MOD400:
! and ax, 3% 'EAX has Year\100, so conditionally equivalent MOD 400
! jnz Done
Julian:
! mov ax, 3%
! and ax, iJahr
! jnz Done
Is29Days:
! mov FebruarTage, 29%
Done:
! pop ax
function = FebruarTage
end function
function pbmain () as long 'put only to demonstrate Tage_im_Februar
local iJahr as integer
iJahr = 2000
print Tage_im_Februar(iJahr); " 2000, divisible by 400, is leap"
iJahr = 2100
print Tage_im_Februar(iJahr); " 2100, divisible by 100 not 400, not leap"
iJahr = 2104
print Tage_im_Februar(iJahr); " 2104, divisible by 4 not 100, is leap"
iJahr = 2103
print Tage_im_Februar(iJahr); " 2103, not divide 4, not leap
print "julian, did not check rule for prior to Gregorian myself"
iJahr = 1204
print Tage_im_Februar(iJahr); " 1204, divisible by 4, is leap
iJahr = 1201
print Tage_im_Februar(iJahr) " 1201, not divisible by 4, not leap
waitkey$
end function

RE: Leap days (or not) in February for PBDOS - Andy Dee - 08.03.2025

Thanks a lot, Dale.

I converted everything to the syntax and rules of PB 3.5 and it workg fine!

Code:
'- 08.03.2025 - Dales work with syntax of PB 3.5 

$Dim All

$Compile exe

Declare _

Function iTage_im_Februar(byval iJahr as integer) as integer

Function iTage_im_Februar(byval iJahr as integer) as integer

  Dim iFebruarTage As Local Integer

  ! push ax

  ! mov iFebruarTage, 28%              ; not a leap year, pre-set to 28

  ! mov ax, 1582%

  ! cmp ax, iJahr

  ! jge Julian

  ! mov ax, 3%

  ! and ax, iJahr                      ; conditionally equivalent MOD 4

  ! jz MOD100                          ; 0 is possible leap year

  ! jmp Done

MOD100:

  ! push bx                            ; for divisor

  ! push dx                            ; for high part of dividend,

                                       ' remainder (MOD)

  ! xor dx, dx

  ! mov ax, iJahr

  ! mov bx, 100%

  ! div bx

  ! cmp dx, 0%                         ; does MOD 100 = 0 ?

  ! pop dx

  ! pop bx

  ! jz MOD400                          ; 0 is possibly not a leap year

  ! jmp Is29Days                       ; non 0 is a leap year

MOD400:

  ! and ax, 3%                         ; AX has Year\100, so conditionally

                                       ' equivalent MOD 400

  ! jnz Done

Julian:

  ! mov ax, 3%

  ! and ax, iJahr

  ! jnz Done

Is29Days:

  ! mov iFebruarTage, 29%

Done:

  ! pop ax

  function = iFebruarTage

End Function

Function pbMain () as long 'put only to demonstrate Tage_im_Februar

  Dim iJahr As Local Integer

  iJahr = 2000

  print iTage_im_Februar(iJahr); " 2000, divisible by 400, is leap"

  iJahr = 2100

  print iTage_im_Februar(iJahr); " 2100, divisible by 100 not 400, not leap"

  iJahr = 2104

  print iTage_im_Februar(iJahr); " 2104, divisible by 4 not 100, is leap"

  iJahr = 2103

  print iTage_im_Februar(iJahr); " 2103, not divide 4, not leap

  print "julian, did not check rule for prior to Gregorian myself"

  iJahr = 1204

  print iTage_im_Februar(iJahr); " 1204, divisible by 4, is leap

  iJahr = 1201

  print iTage_im_Februar(iJahr)  " 1201, not divisible by 4, not leap

end function

pbMain

The output is the following:

Code:
29  2000, divisible by 400, is leap

 28  2100, divisible by 100 not 400, not leap

 29  2104, divisible by 4 not 100, is leap

 28  2103, not divide 4, not leap

julian, did not check rule for prior to Gregorian myself

 29  1204, divisible by 4, is leap

 28  1201, not divisible by 4, not leap

This helps me a lot, Dale!

RE: Leap days (or not) in February for PBDOS - Andy Dee - 09.03.2025

Bit optimized routines, regarding to your thoughts:

Code:
Declare _

FUNCTION iFeb(ByVal iJahr As Integer)        As Integer

'----------------------------------------------------------------------------

' Optimize the BASIC code for speed

FUNCTION iFeb(ByVal iJahr As Integer) Public As Integer

  '--------------------------------------------------------------------------

  ' Use a temporary variable to reduce calculationtime

  Dim iJahriDiv4 As Local Integer

  '--------------------------------------------------------------------------

  ' Optimize the BASIC's mathematics - thanks to Dale Yarker

  $If 0

    Arithmetic ops by powers of 2 can be done a bit faster by not using

    *, \ or MOD (likely you know of 2 them). For example:

    X * 4 can be SHIFT LEFT X, 2

    X \ 4 can be SHIFT RIGHT X, 2

    X MOD 4 can be X AND 3

    Those can be done in BASIC code so difference between BASIC and ASM

    timing may be made smaller.

    MOD is a divide with the remainder returned instead of the quotient.

    Your BASIC code can be quicker with MOD 100 and MOD 400 than 2 divides

    by 100 and 2 divides by 400.

    In ASM the DIV instruction returns both quotient and remainder. 100 is

    not a power 0f 2, so had to divide by 100. The remainder used as

    MOD 100 and quotient used with X AND as MOD 400.

    The 1,15 times faster for 1204 and 1201 shows how small the difference

    is between MOD and AND.)

  $EndIf

  iJahriDiv4  = iJahr

  SHIFT RIGHT    iJahriDiv4, 2%

  if iJahr < 1582% then

    Function = 29% _

            + ((iJahr AND 3)<>0%)

  else

    Function = 29% _

            + ( _

                (iJahr/4%) <> iJahriDiv4  _

                OR _        _

                (iJahr/100%)=(iJahr\100%) _

                AND _

                (iJahr/400%)<>(iJahr\400%) _

              )

  end if

END FUNCTION

Declare _

Function iTage_im_Februar(byval iJahr as integer) as integer

'----------------------------------------------------------------------------

' Optimize the BASIC code for speed

Function iTage_im_Februar(byval iJahr as integer) as integer

  Dim iFebruarTage As Local Integer

  ! mov iFebruarTage, 28%              ; not a leap year, pre-set to 28

  ! mov ax, 1582%

  ! cmp ax, iJahr

  ! jge Julian

  ! mov ax, 3%

  ! and ax, iJahr                      ; conditionally equivalent MOD 4

  ! jz MOD100                          ; 0 is possible leap year

  ! jmp Done

MOD100:

                                      ' remainder (MOD)

  ! xor dx, dx

  ! mov ax, iJahr

  ! mov bx, 100%

  ! div bx

  ! cmp dx, 0%                        ; does MOD 100 = 0 ?

  ! jz MOD400                          ; 0 is possibly not a leap year

  ! jmp Is29Days                      ; non 0 is a leap year

MOD400:

  ! and ax, 3%                        ; AX has Year\100, so conditionally

                                      ' equivalent MOD 400

  ! jnz Done

Julian:

  ! mov ax, 3%

  ! and ax, iJahr

  ! jnz Done

Is29Days:

  ! mov iFebruarTage, 29%

Done:

  function = iFebruarTage

End Function