为什么在此绑定函数中不需要上载? [英] Why upcast is not necessary in this let bound function?

问题描述

根据 MSDN

将参数传递给方法时，将自动应用向上转换 在对象类型上.但是，对于模块中的绑定函数， 除非参数类型声明为 灵活的类型.

Upcasting is applied automatically when you pass arguments to methods on an object type. However, for let-bound functions in a module, upcasting is not automatic, unless the parameter type is declared as a flexible type.

但是

type C() =
    member this.T() = ()
type D() =
    inherit C()

let myfun (x: C)=
    ()

let d = D()

myfun d

我根本不需要上流.

推荐答案

看看 这意味着其推断类型在参数位置包括未密封类型的F#函数可以在调用时传递为子类型，而无需显式上载.例如:

(该示例显示的内容几乎与您的相同)

(The example showed there is almost the same as yours)

还请注意，对于绑定函数，除非在某些情况下(例如，当参数组成时，如果您的函数是:

Also note that for let bound functions upcasting is automatically except in some cases, for example when the argument is composed, if your function was:

let myfun (x: C option)=
    ()

上传不再自动.

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