Question

我在Swift中找到了你可以存储静态方法的引用而不实际执行它：

class StringMama {
    static func returnString(s:String)->String {
      return s
    }
}

var stored = StringMama.returnString
print(stored.self) // (Function)
print(stored(s:"Hello!")) // for some reason it doesn't work
print(stored("Hello!")) // it works

现在，我希望在函数体中传递stored作为函数的参数，以便稍后执行该函数。这可能吗？怎么样？我找不到办法。任何帮助表示赞赏

Answer 1

    printUsingReference(stored, "the")


static func printUsingReference(_ stored: (_ s: String) -> String, _ content: String) {
    print(stored(content))

}

Answer 2

你可以这样做：

func Foo(param: (String) -> String) {
    print(param("Foo called"))
}

// Pass it your stored var
Foo(param: stored)

Answer 3

为什么不使用封口？存储对变量的引用并从中获取值并传入函数。

var someValue: (String) -> String = { stringValue in
  return stringValue
}

func someFunction(value: String) {
 print(value)
}

//Now pass closure reference into one variable 
let value = someValue
//Using in that function like below.
someFunction(value: value("test"))

将静态方法的引用传递给函数

3 个答案: