我尝试应用Zeller的收敛简化方法从用户输入日期中获取日期名称。
的简化算法
Zeller's Congruence
variable year 2 allot
variable day 2 allot
variable month 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 month 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)
month @ case
1 of
month 13 !
year @ 1- !
endof
2 of
month 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 @ month @ 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
语法似乎还可以，但方法或错误/失败的函数可能是根本原因。
输入很好，但随机获得的日期并不好(即使是同一天(。
所以我可能没能做点什么&在这里，我取消了代码的优化，试图调试它，但我还没有找到原因。
下面是一个执行示例：
插入日期：
gforth zellersconvergence_bugged.fs
redefined k redefined j
Insert a decomposed date:
Century ? 20
Year ? 21
Mounth ? 6
Day ? 8
Tuesday !
gforth zellersconvergence_bugged.fs
redefined k redefined j
Insert a decomposed date:
Century ? 20
Year ? 21
Mounth ? 6
Day ? 8
Saturday !
gforth zellersconvergence_bugged.fs
redefined k redefined j
Insert a decomposed date:
Century ? 20
Year ? 21
Mounth ? 6
Day ? 8
Monday !
这会是堆栈问题吗？
这会是一个方法问题吗？
这可能是算法内部被误解的东西吗？
您需要两次从年份变量year @ 100 ...中获取数据。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。作为变量的j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！！
我会这样实现它。然后，我可以使用堆栈输入在控制台中运行这些单词，看看发生了什么来帮助调试。
: 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 !