我开始用haskell编程.我正在开发的程序只是将一个列表的总和与两个元素相加,例如:
[("book",10),("cookies",2),("icecream",5)]
这应该返回"17".这是我的代码:
total [] = [] total ([("c",e)]:y) = total y ++ [e]
但是在GHCi中运行它会给我这个错误:
:80:8: Couldn't match expected type `[([Char], a0)]' with actual type `([Char], t0)' In the expression: ("livro", 10) In the first argument of `total', namely `[("livro", 10), ("bolachas", 2), ("gelado", 5)]' In the expression: total [("livro", 10), ("bolachas", 2), ("gelado", 5)] :80:21: Couldn't match expected type `[([Char], a0)]' with actual type `([Char], t1)' In the expression: ("bolachas", 2) In the first argument of `total', namely `[("livro", 10), ("bolachas", 2), ("gelado", 5)]' In the expression: total [("livro", 10), ("bolachas", 2), ("gelado", 5)] :80:36: Couldn't match expected type `[([Char], a0)]' with actual type `([Char], t2)' In the expression: ("gelado", 5) In the first argument of `total', namely `[("livro", 10), ("bolachas", 2), ("gelado", 5)]' In the expression: total [("livro", 10), ("bolachas", 2), ("gelado", 5)]
这可能很简单,但作为初学者,我无法解决这个问题.
在线:
total ([("c",e)]:y) = total y ++ [e]
在([("c",e)]:y)
没有做你想要的.它匹配一个非空列表,其中第一个元素也是一个列表(因为[...]
),并且该子列表中只有一个元素,即第一个元素等于的对"c"
.为了匹配你想要的东西,你需要写:
total ((c,e):y) = total y ++ [e]
但是,这仍然无法执行您想要的操作,因为它构造了输入列表中所有e
值的列表.要将它们相加,您需要:
total [] = 0 total ((c,e):y) = total y + e