将静态方法的引用传递给函数

Question

我在 Swift 中发现您可以存储静态方法的引用而无需实际执行它：

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，我想将 stored 作为函数的参数传递，以便稍后在函数体中执行该函数。这可能吗？如何？我找不到办法。感谢任何帮助

Answer 1

    printUsingReference(stored, "the")


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

}

Answer 2

您可以这样做：

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

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

Answer 3

为什么不使用闭包？存储它对变量的引用并从中获取值并传入您的函数。

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

Pass reference of static method to a function

static-methods

function

parameter-passing

swift