包装函数

Question

在 swift 函数中，不提供名义类型的常规功能。这意味着唯一的解决方案是将它们包装到某个对象中。在struct，例如：

struct Functor<Input, Output> {
    //assuming only fuctions with one argument
    let function: (Input) -> Output
    let inputType: Input.Type
    let outputType: Output.Type
    let input: Input
    let output: Output

    init<Input, Output>(_ function: @escaping (Input) -> Output) {
        self.function = function //there is already a compilation error
        self.input = Input //???? And I have not a clue on how to store argument as instance
    }
}

编译器说：

error: cannot assign value of type '(Input) -> Output' to type '(Input) -> Output'

有没有办法让它成为可能，还是由于类型系统设计缺陷而无法实现？

Answer 1

当您在泛型类型中声明泛型方法时，内部泛型参数（在方法上）会影响外部泛型参数。

你不应该让你的初始化器通用：

struct Functor<Input, Output> {
    //assuming only fuctions with one argument
    let function: (Input) -> Output
    let inputType: Input.Type
    let outputType: Output.Type
    //### I do not understand how you want to use the followings...
//    let input: Input
//    let output: Output

    init(_ function: @escaping (Input) -> Output) { //### Do not put inner generic parameters
        self.function = function
        //### I do not understand how you want to use the followings...
        self.inputType = Input.self
        self.outputType = Output.self
    }
}

我不明白你真正想对第二个错误行做什么，所以请添加评论或更新你的问题以澄清这一点。