如何将自定义类型调用函数泛化为抽象类型? [英] How can I generalize a custom type call function to an abstract type?

问题描述

我有以下模拟设置，具有抽象类型，具体类型作为子类型以及带有两个参数的函数f，第一个是Letter.

I have the following mock setup, with an abstract type, concrete types as subtypes, and a function f which takes two arguments, the first being a Letter.

abstract type Letter end

struct A <: Letter end
struct B <: Letter end

f(::A, x) = ('a', x)
f(::B, x) = ('b', x)

a = A()
b = B()

我想为Letter子类型定义一个自定义调用函数，该函数简单地调用f.

I would like to define a custom call function for Letter subtypes which simply calls f.

(t::A)(x) = f(t, x)
(t::B)(x) = f(t, x)

虽然可行，但似乎很多余，尤其是考虑到Letter的子类型可能更多.我的尝试如下，但似乎都无效.

While this works, it seems quite redundant, especially considering that there could be many more subtypes of Letter. My attempts are as follows, but neither seem to work.

julia> (t::Letter)(x) = f(t, x)
ERROR: cannot add methods to an abstract type

julia> (t::T)(x) where T <: Letter = f(t, x)
ERROR: function type in method definition is not a type

如何概括调用函数以匹配Letter的任何(具体)子类型?

How can I generalize a call function to match any (concrete) subtype of Letter?

推荐答案

以Dan的回答为基础，元编程似乎是必经之路

Building on Dan's answer, metaprogramming seems to be the way to go

for T in Symbol.(subtypes(Letter))
    @eval (t::$T)(x) = f(t, x)
end

为每种类型生成函数.

或者:

for T in Symbol.(subtypes(Letter))
    c = Char(lowercase(first(String(T))))
    @eval f(::$T, x) = ($c, x)
    @eval (t::$T)(x) = f(t, x)
end

但是，这种将结构/子类型用作枚举的方法是

However, this use of structs/subtypes as enums is discouraged

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