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，有一个抽象类型，具体类型作为子类型，以及一个函数 f，它有两个参数，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，它只调用 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)

虽然这行得通，但似乎很多余，尤其是考虑到 Letter 可能有更多的子类型.我的尝试如下，但似乎都不起作用.

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

基于 Dan 的回答，元编程似乎是要走的路

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

按类型生成函数.

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

然而，这种使用结构体/子类型作为枚举是不鼓励

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

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