如何在不传入值的情况下调用带参数的函数？ [迅捷游乐场]

Question

Swift Playground 提供了以下代码。不传入参数如何调用 speakText(graphic: )？ （显然图形已经放在另一段代码中了）

// Speak the text of graphic.
func speakText(graphic: Graphic) {
    speak(graphic.text)
}
func addGreeting(touch: Touch) {
    if touch.previousPlaceDistance < 60 { return }
    let greetings = ["howdy!", "hello", "hi", "ciao", "yo!", "hey!",       "what’s up?"]
    let greeting = greetings.randomItem
    let graphic = Graphic(text: greeting)
    graphic.textColor = #colorLiteral(red: 0.9686274529, green: 0.78039217, blue: 0.3450980484, alpha: 1)
    graphic.fontName = .chalkduster
    scene.place(graphic, at: touch.position)
    graphic.rotation = randomDouble(from: -30, to: 30)
}
// Create and add Speak tool.
let speakTool = Tool(name: "Speak", emojiIcon: "????")
speakTool.onGraphicTouched = speakText(graphic: )
scene.tools.append(speakTool)

Answer 1

speakTool 是Tool 类型，它有一个onGraphicTouched 类型的属性(Graphic) -> ()，它是一个函数/闭包，它接受Graphic 作为输入并且什么都不返回（Void 或@987654329 @)。

speakText(graphic:) 是指向上面定义的函数的函数指针。请注意，该函数具有所需的签名；它需要 Graphic 并且什么都不返回。

所以speakTool.onGraphicTouched = speakText(graphic: ) 将指向函数的指针分配给onGraphicTouched，当触摸图形时，speakTool 将调用onGraphicTouched(someGraphic)，而speakText(graphic: someGraphic) 将调用。

您可以在Apple's Swift Guide.中的函数类型部分阅读更多相关信息