输入奥秘。为什么这个代码编译？ [英] Type mysteries. Why does this code compile?

问题描述

此代码无法编译：

  default（）
 
f :: RealFloat a => ; a 
 f = 1.0 
 
 g :: RealFloat a => a 
 g = 1.0 
 
 h :: Bool 
 h = f < g  - 错误。暧昧。 
  

这是预料之中的，因为它不明确。这两种可能性是 Float 和 Double ，编译器不知道哪个< 选择。

然而，这段代码并不会编译：

  default（）
 
f :: RealFloat a => a 
 f = 1.0 
 
 g :: RealFloat a => a 
 g = 1.0 
 
 h :: RealFloat a => a 
 h = f + g  - 为什么编译？ 
  

为什么？为什么Haskell在这里遇到困惑，与上面的例子类似，需要选择 + （for Float c $ c> h 也是具有多态类型。因此，使用 + 的类型不是不明确的;它只是还没有被选中。

使用 将决定选择哪种类型的 + （并且不同的使用地点可以做出不同的选择）。 h 的用户可以要求它提供他们所需的任何 RealFloat 类型; f 和 g 也可以提供任何 RealFloat 类型，所以 h 只会询问他们的用户所要求的类型。

This code does not compile:

default ()

f :: RealFloat a => a
f = 1.0

g :: RealFloat a => a
g = 1.0

h :: Bool
h = f < g --Error. Ambiguous.

This is expected because it's ambiguous. The two possibilities are Float and Double and the compiler doesn't know which < to pick.

However, this code does compile:

default ()

f :: RealFloat a => a
f = 1.0

g :: RealFloat a => a
g = 1.0

h :: RealFloat a => a
h = f + g --Why does this compile?

Why? Why isn't Haskell confused here, in a similar manner as in the example above, about which + to pick (for Float or Double)?

解决方案

In your second example h also has a polymorphic type. So the type at which + is used isn't ambiguous; it just hasn't been chosen yet.

The context where h is used will determine which type's + is chosen (and different use-sites can make different choices). The user of h can ask it to provide any RealFloat type they please; f and g can also provide any RealFloat type, so h will just ask them for exactly the type its user is asking for.

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