如何解决gforth 从用户指定的日期获取日期名称
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。
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
输入很好,但随机获得的那一天不是好的(即使是同一天)
所以我可能没有做某事&在这里我未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? 可能是方法问题? 这可能是算法本身内部被误解的事情?
谢谢
解决方法
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,我可以使用堆栈输入在控制台中运行这些单词,以查看正在发生的情况以帮助调试。
: 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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