1
我有這樣的功能:如何存儲這種返回類型:[(x,y)]?
example :: [Char] -> [Char]
example myString = ...................
where
pat = "something"
returnList = myString =~ pat :: [(MatchOffset,MatchLength)]
我的問題是,我不知道如何存儲的值,我回去與調用myString =~ pat :: [(MatchOffset,MatchLength)]
我不能只將其存儲在一個單一的變量名稱,因爲我已經在這裏完成了,但我不確定我是如何存儲它的。
目前,它給出了這樣的錯誤:
No instance for (RegexContext
Regex [Char] [(MatchOffset, MatchLength)])
arising from a use of `=~'
Possible fix:
add an instance declaration for
(RegexContext Regex [Char] [(MatchOffset, MatchLength)])
In the expression: myString =~ pat :: [(MatchOffset, MatchLength)]
In an equation for `returnList':
returnList = myString =~ pat :: [(MatchOffset, MatchLength)]
In an equation for `example':
example myString
= ....................
where
pat = "something"
returnList = myString =~ pat :: [(MatchOffset, MatchLength)]
完美,謝謝! –