Swift:调用返回类型不明确的重载方法 [英] Swift: call overloaded method with ambiguous return type

问题描述

假设我有以下代码:

 公共类 A {}公共类 B : A {}公共类 Foo {public func bar() ->一种 {print("秘密隐藏方法")返回 A()}public func bar() ->乙{打印(易于访问的方法")返回 B()}}

问题是尝试调用它们会导致歧义.您可以调用 bar() ->B 以 Foo().bar() 作为 B，但所有的 Foo().bar(), Foo().bar() 作为 A，并且 let a: A = Foo().bar() 产生 Ambiguous use of 'bar()' 编译器错误.>

底线:我如何调用 bar() ->A? 是否有一些棘手的语法?有必要反思吗?反射足够吗?

解决方案

编译器需要能够确定您想要执行 bar() 的哪个变体.您可以明确要执行的 func 类型，并利用柯里化来访问给定实例的正确实现.

let foo = Foo()让 a: ()->A = Foo.bar(foo)让 b: ()->B = Foo.bar(foo)一种()乙()

上面的代码会打印:

秘密隐藏方法易于访问的方法

Suppose I have the following code:

    public class A {}
    public class B : A {}
    public class Foo {
        public func bar() -> A {
            print("secret hidden method")
            return A()
        }

        public func bar() -> B {
            print("easily accessible method")
            return B()
        }
    }

The problem is try to call them leads to ambiguity. You can call bar() -> B with Foo().bar() as B, but all of Foo().bar(), Foo().bar() as A, and let a: A = Foo().bar() yield an Ambiguous use of 'bar()' compiler error.

Bottom line: How do I call bar() -> A? Is there some tricky syntax for it? Is reflection necessary? Is reflection sufficient?

解决方案

The compiler needs to be able to determine which variation of bar() you want to execute. You can be explicit about the type of func you want to execute, and take advantage of currying to access the correct implementation for a given instance.

let foo = Foo()
let a: ()->A = Foo.bar(foo)
let b: ()->B = Foo.bar(foo)
a()
b()

The above code will print:

secret hidden method
easily accessible method

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