如果我查看它的文档space
建议使用void spaceChar
.
但是,如果我真的尝试:
x :: Parser ()
x = void spaceChar
我明白了
* Couldn't match type `Token s0' with `Char'
arising from a use of `spaceChar'
The type variable `s0' is ambiguous
* In the first argument of `void', namely `spaceChar'
In the expression: void spaceChar
In an equation for `x': x = void spaceChar
我认为表达式是正确的,但我需要做一些事情来说服类型检查器。我怎样才能让它工作?