该字符串如下所示:
",w84,w41,w56,w170,w56,w41,w84,/,,w24,w40,w17,w40,w48,,/ ,,,w16,w16,w16,,,/,,,,,,,,/,,,,,,,,/,,,,,,,,/,,,b1,b1,b1,,,/ ,,b3,b130,b17,b130,b129,,/,b69,b146,b131,b170,b131,b146,b69,"但应该是这样的
[[Empty,Piece White 84,Piece White 41,Piece White 56,Piece White 170,Piece White 56,Piece White 41,Piece White 84,Empty],[Empty,Empty,Piece White 24,Piece White 40,Piece White 17,Piece White 40,Piece White 48,Empty,Empty],[Empty,Empty,Empty,Piece White 16,Piece White 16,Piece White 16,Empty,Empty,Empty],[Empty,Empty,Empty,Empty,Empty,Empty,Empty,Empty,Empty],[Empty,Empty,Empty,Empty,Empty,Empty,Empty,Empty,Empty],[Empty,Empty,Empty,Empty,Empty,Empty,Empty,Empty,Empty],[Empty,Empty,Empty,Piece Black 1,Piece Black 1,Piece Black 1,Empty,Empty,Empty],[Empty,Empty,Piece Black 3,Piece Black 130,Piece Black 17,Piece Black 130,Piece Black 129,Empty,Empty],[Empty,Piece Black 69,Piece Black 146,Piece Black 131,Piece Black 170,Piece Black 131,Piece Black 146,Piece Black 69,Empty]]我的代码创建的列表如下所示:
[“空”、“白”、“空、空、白24、白40、白17、白40、白48、空、空”、“空、空”、“空”
data Player = Black | White deriving Show
data Cell = Piece Player Int | Empty deriving Show
data Pos = Pos { col :: Char, row :: Int } deriving Show
type Board = [[Cell]]我有这些数据类型。
我几乎完成了这个任务,我所需要的就是去掉引号。
到目前为止,这是我的代码:
buildBoard x = rec(help3(wop(helper (replaceO x))))
wop (x:xs) = splitOn "/" (x:xs)
help3 (x:xs) = map (\x -> [x])(x:xs)
rec (x:xs) = map(\x -> [recH(x)])(x:xs)
recH (x:xs) = checkComma(x)
helper (x:y:xs)
|x == ',' && y == ',' = x:'E':'m':'p':'t':'y':helper(y:xs)
|otherwise = x:helper (y:xs)
helper [] = []
helper [x] = [x]
checkComma (x:xs) = if head (x:xs) == ',' then checkComma('E':'m':'p':'t':'y':',':xs) else if last (x:xs) == ',' then reverse(turnAr(reverse(x:xs))) else (x:xs)
turnAr (x:xs) = 'y':'t':'p':'m':'E':',':xs
replaceO [] = []
replaceO (x:xs) =
if x == 'w'
then 'P':'i':'e':'c':'e':' ':'W':'h':'i':'t':'e':' ': replaceO xs
else if x == 'b'
then 'P':'i':'e':'c':'e':' ':'B':'l':'a':'c':'k':' ': replaceO xs
else if x == 'E'
then 'E':'m':'p':'t':'y':' ': replaceO xs
else x : replaceO xs发布于 2022-12-04 16:10:15
仅仅去掉引号不是一件容易的事。
我认为您必须对代码进行一些重大更改。我建议先执行splitOn "/",然后再执行splitOn ",",它生成一个字符串列表,每个字符串代表一个单元格。
然后,您可以很容易地编写一个函数parseCell :: String -> Cell来解析这些内部单元格。这个函数将有点像您的replaceO函数,但它也应该处理所有空单元格并实际解析整数(您可以使用read函数)。
https://stackoverflow.com/questions/74677284
复制相似问题