Question

我试图编写3个函数。

ghci> let f = floor . (logBase 2) . length

但是，我不理解这个编译时错误。

<interactive>:47:9:
    No instance for (RealFrac Int) arising from a use of `floor'
    Possible fix: add an instance declaration for (RealFrac Int)
    In the first argument of `(.)', namely `floor'
    In the expression: floor . (logBase 2) . length
    In an equation for `f': f = floor . (logBase 2) . length

<interactive>:47:18:
    No instance for (Floating Int) arising from a use of `logBase'
    Possible fix: add an instance declaration for (Floating Int)
    In the first argument of `(.)', namely `(logBase 2)'
    In the second argument of `(.)', namely `(logBase 2) . length'
    In the expression: floor . (logBase 2) . length

在此输出中：

    No instance for (RealFrac Int) arising from a use of `floor'
    Possible fix: add an instance declaration for (RealFrac Int)

这是否意味着我需要明确指定类型（通过:: some_type_here）而不是依靠Haskell来推断类型？

Answer 1

那是因为logbase的类型是：

λ> :t logbase
logBase :: Floating a => a -> a -> a

因此，他们接受的类型是Floating类型类的实例，即。 Float和Double数据类型。因此，如果您进行了正确的类型转换，它应该可以工作：

λ> let f = floor . (logBase 2) . fromIntegral . length

RealFrac和Floating Int的函数组合

1 个答案: