gforth 从用户指定的日期获取日名称

Question

我尝试应用 zeller 的收敛简化方法从用户输入的日期中获取日期名称。

来自

的简化算法

\ Zeller's Congruence
variable year       2   allot
variable day        2   allot
variable mounth     2   allot
variable century    2   allot
variable daynumber  1   allot
variable k          2   allot
variable j          2   allot

\ read keyboard word input
: input$ ( n -- addr n )
        pad swap accept
        pad swap
;

\ check input type
: input# ( -- u true | false )
        0. 16 input$ dup >R
        >number nip nip
        R> <> dup 0 = if
                nip
        then
;

\ get all year mounth and day to check
: readyear
        CR ." Year    ? "
        input# if
                year !
        else
                cr ." Must be a number" cr
                bye
        then
        year @ dup >r 99 > r> 1 < or if \ more forth way to write it
                cr ."  Must be lower than 99 and gregorian date ( so also over 1752 September 2cd)" cr
                bye
        then
;

: readday
        CR ." Day     ? "
        input# if
                day !
        else
                cr ." Must be a number" cr
                bye
        then
        day @ dup >r 31 > r> 1 < or if
                cr ."  Must be between 1 and 31" cr ( is user stupid ? )
                bye
        then
;

: ?adaptday
                \ NOTE: In this algorithm January and February are
                \      counted as months 13 and 14 of the previous
                \      year.E.g. if it is 2 February 2010, the
                \      algorithm counts the date as the second day
                \      of the fourteenth month of 2009 (02/14/2009
                \      in DD/MM/YYYY format)
        mounth @ case
            1 of
                mounth 13 !
                year @ 1- !
            endof
            2 of
                mounth 14 !
                year @ 1- !
            endof
        endcase
        \                      13(m+1)          K       J
        \ daynumber = ( day + (-------) + k + (---) + (---) + 5j ) %7
        \                         5             4       4
        year 100 mod k !
        year 100 / j !
        day @ mounth @ 1 + 13 * 5 / +       \ day + ((13*(m-1))/5)
        k @ +                               \ day + ((13*(m-1))/5) + k
        k @ 4 / +                           \ day + ((13*(m-1))/5) + k + k/4
        J @ 4 / +                           \ day + ((13*(m-1))/5) + k + k/4 + J/4
        J @ 5 * +                           \ day + ((13*(m-1))/5) + k + k/4 + J/4 + 5J
        7 mod daynumber !                   \ (day + ((13*(m-1))/5) + k + k/4 + J/4 + 5J) %7
        \ 1 line for each sub calculation just for better mathematical reading
        daynumber @ case
            0 of cr ."      Saturday       !" cr bye endof
            1 of cr ."      Sunday         !" cr bye endof
            2 of cr ."      Monday         !" cr bye endof
            3 of cr ."      Tuesday        !" cr bye endof
            4 of cr ."      Wednesday      !" cr bye endof
            5 of cr ."      Thursday       !" cr bye endof
            6 of cr ."      Friday         !" cr bye endof
        endcase
;

\ main function
: main
        page
        cr
        >readvars
        ?adaptday
        cr cr
        bye
;

main

语法似乎没问题，但方法或 buggy/failed 函数可能是根本原因。

输入很好，但随机获得的日期不是很好（即使是同一日期）

所以我可能没有做某事 & 在这里我取消优化代码以尝试调试它，但我还没有找到原因。

这是一个执行示例：

插入日期：

francois@zaphod:~/GITLAB/dev/dev_gforth_calendar$ gforth zellersconvergence_bugged.fs
redefined k  redefined j


Insert a decomposed date :

Century ? 20
Year    ? 21
Mounth  ? 6
Day     ? 8

     Tuesday        !
francois@zaphod:~/GITLAB/dev/dev_gforth_calendar$ gforth zellersconvergence_bugged.fs
redefined k  redefined j


Insert a decomposed date :

Century ? 20
Year    ? 21
Mounth  ? 6
Day     ? 8

     Saturday       !
francois@zaphod:~/GITLAB/dev/dev_gforth_calendar$ gforth zellersconvergence_bugged.fs
redefined k  redefined j


Insert a decomposed date :

Century ? 20
Year    ? 21
Mounth  ? 6
Day     ? 8

     Monday         !
francois@zaphod:~/GITLAB/dev/dev_gforth_calendar$

这可能是堆栈问题？这可能是方法问题？算法本身可能是一个被误解的东西？

谢谢

Answer 1

您需要从 year 变量中获取数据两次，year @ 100 ...。我认为在那之后 ?adaptday 会起作用。第四个单词 within \ n lo hi -- flag ; flag is True if lo <= n < hi 用于检查范围内的数字，

在 Forth 中使用这么多变量是不寻常的。这些值通常存储在堆栈中。 j 作为变量可以覆盖用作外部 do 循环计数器的 j。我也看到 k 用于下一个外循环！！

我会像这样实现它。然后我可以运行控制台中的单词和堆栈输入来查看正在发生的事情以帮助调试。

: century-ix  \ c -- days-ix
  dup 4/ swap 5 * + 
;
: year-ix   \ yy -- days-ix
  dup 4/ +  
;
: month-ix  \ mm - days-ix
  1+ 13 * 5 / 
;
: weekday \ dd mm yyyy -- dow
  over 3 < if
    swap 12 + swap 1-  \ adjusts Jan and Feb to be month 13 or 14 of previous year.
  then 
  100 /mod  ( dd mm yy cc )
  century-ix            ( dd mm yy days )
  swap year-ix +        ( dd mm days )
  swap month-ix + +     \ Calculate for months and days
  7 mod 
; 

: weekday.  \ n -- ;  -- Prints weekday in English
            \  Too useful to hide in another definition.
  case
        0 of cr ."      Saturday       !" cr  endof
        1 of cr ."      Sunday         !" cr  endof
        2 of cr ."      Monday         !" cr  endof
        3 of cr ."      Tuesday        !" cr  endof
        4 of cr ."      Wednesday      !" cr  endof
        5 of cr ."      Thursday       !" cr  endof
        6 of cr ."      Friday         !" cr  endof
  endcase
; 

8 6 2021 weekday weekday. 
     Tuesday        !

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