通过具有不同参数数量的方程来定义函数 [英] Defining a function by equations with different number of arguments

问题描述

我今天注意到这样一个定义

  safeDivide x 0 = x 
 safeDivide =（/）
  

是不可能的。我只是好奇这背后的（好）原因是什么。必须有一个非常好的（这是Haskell毕竟:)）。

注意：我没有看上述代码的替代实现的建议，这是一个简单的例子以证明我的观点。

解决方案

我认为这主要是为了保持一致性，因此所有的子句都可以用相同的方式读取，说话;即每个RHS在功能类型中处于相同的位置。我认为如果你允许的话，也会掩盖一些愚蠢的错误。

还有一个轻微的语义怪癖：说编译器填充这些子句以得到相同的数字作为其他条款的模式;即你的例子将变成

  safeDivide x 0 = x 
 safeDivide xy =（/）xy 
  

现在考虑第二行是否为 safeDivide = undefined ;在没有前面的子句的情况下， safeDivide 会是⊥，但是由于这里执行的eta扩展， \xy - >如果y == 0那么x else⊥ - 所以 safeDivide = undefined 实际上并没有定义 safeDivide 成为⊥！这似乎令人困惑，足以证明禁止此类条款，国际海事组织。

I noticed today that such a definition

safeDivide x 0 = x
safeDivide = (/)

is not possible. I am just curious what the (good) reason behind this is. There must be a very good one (it's Haskell after all :)).

Note: I am not looking suggestions for alternative implementations to the code above, it's a simple example to demonstrate my point.

解决方案

I think it's mainly for consistency so that all clauses can be read in the same manner, so to speak; i.e. every RHS is at the same position in the type of the function. I think would mask quite a few silly errors if you allowed this, too.

There's also a slight semantic quirk: say the compiler padded out such clauses to have the same number of patterns as the other clauses; i.e. your example would become

safeDivide x 0 = x
safeDivide x y = (/) x y

Now consider if the second line had instead been safeDivide = undefined; in the absence of the previous clause, safeDivide would be ⊥, but thanks to the eta-expansion performed here, it's \x y -> if y == 0 then x else ⊥ — so safeDivide = undefined does not actually define safeDivide to be ⊥! This seems confusing enough to justify banning such clauses, IMO.

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