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

问题描述

我今天注意到这样的定义

I noticed today that such a definition

safeDivide x 0 = x
safeDivide = (/)

是不可能的.我只是好奇这背后的(好的)原因是什么.一定有一个非常好的(毕竟是 Haskell :)).

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 :)).

注意:我不是在寻找上述代码的替代实现建议，这是一个简单的例子来证明我的观点.

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

推荐答案

我认为这主要是为了一致性，以便所有子句可以以相同的方式阅读，可以这么说；即每个 RHS 在函数类型中的相同位置.如果您也允许这样做，我认为会掩盖不少愚蠢的错误.

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.

还有一个轻微的语义怪癖:假设编译器填充了这些子句，使其具有与其他子句相同数量的模式；即你的例子会变成

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

现在考虑第二行是否是 safeDivide = undefined;如果没有前面的条款，safeDivide 将是 ⊥，但由于这里执行的 eta 扩展，它是 xy ->if y == 0 then x else ⊥ — 所以 safeDivide = undefined 实际上并没有将 safeDivide 定义为 ⊥！IMO，这似乎足以证明禁止此类条款的合理性.

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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