haskell - 如何在 Haskell 中进行类型转换？

Question

我在文件 main.hs 中有以下代码:

teste :: Integral a => a -> a
teste n = truncate (sqrt n) + mod n 2

当我尝试加载它时，我从 ghci 收到以下错误:

Prelude> :l main.hs 
[1 of 1] Compiling Main             ( main.hs, interpreted )

main.hs:12:11:
    Could not deduce (RealFrac a) arising from a use of ‘truncate’
    from the context (Integral a)
      bound by the type signature for teste :: Integral a => a -> a
      at main.hs:11:10-29
    Possible fix:
      add (RealFrac a) to the context of
        the type signature for teste :: Integral a => a -> a
    In the first argument of ‘(+)’, namely ‘truncate (sqrt n)’
    In the expression: truncate (sqrt n) + mod n 2
    In an equation for ‘teste’: teste n = truncate (sqrt n) + mod n 2

main.hs:12:21:
    Could not deduce (Floating a) arising from a use of ‘sqrt’
    from the context (Integral a)
      bound by the type signature for teste :: Integral a => a -> a
      at main.hs:11:10-29
    Possible fix:
      add (Floating a) to the context of
        the type signature for teste :: Integral a => a -> a
    In the first argument of ‘truncate’, namely ‘(sqrt n)’
    In the first argument of ‘(+)’, namely ‘truncate (sqrt n)’
    In the expression: truncate (sqrt n) + mod n 2
Failed, modules loaded: none.

但是当我在交互模式下运行相同的代码时，它工作正常:

Prelude> truncate (sqrt 5) + mod 5 2
3

Answer 1

在您的来电 truncate (sqrt 5) + mod 5 2 , 5 s 有不同的类型。确实是5在 sqrt 5应该有类型 Floating a => a ，而 5在 mod 5 2有类型 Integral b => b .尽管严格来说可以在 Haskell 中构造一个属于这两个类型家族的类型，但从概念上讲，一个类型同时是 Integral 是很奇怪的。和 Floating ，它也仅适用于此类类型，使其不太有用。因此，我们可以将签名更改为:

teste :: (Integral a, Floating a, RealFrac a) => a -> a
teste n = truncate (sqrt n) + mod n 2

但如前所述，这不是很有用。

您可以使用 fromIntegral :: (Integral a, Num b) => a -> b 在这里从 Integral 转换输入任意 Num类型，如:

teste :: Integral a => a -> a
teste n = truncate (sqrt (fromIntegral n)) + mod n 2

例如:

Prelude> teste 5
3