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中收到以下错误：

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，它运行良好：

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

Answer 1

在您的通话truncate (sqrt 5) + mod 5 2中，5的类型不同。实际上，{{1}中的5应该具有类型sqrt 5，而Floating a => a中的5具有类型mod 5 2。尽管严格说来，可以在Haskell中构造属于两个类型族的成员的类型，但是从概念上说，类型既是Integral b => b也是Integral，它也仅适用于此类类型，从而使其用处不大。因此，我们可以将签名更改为：

Floating

但是如前所述，这不是非常有用。

您可以在此处使用fromIntegral :: (Integral a, Num b) => a -> b从teste :: (Integral a, Floating a, RealFrac a) => a -> a teste n = truncate (sqrt n) + mod n 2类型转换为任何Integral类型，例如：

Num

例如：

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

如何在Haskell中进行类型转换？

1 个答案: