swift - 如何在不传递值的情况下调用带参数的函数？ [ swift Playground ]

Question

Swift Playground 提供了以下代码。如何在不传递参数的情况下调用 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 属于 Tool 类型，它具有 onGraphicTouched 类型的属性 (Graphic) -> () 这是一个函数/闭包，它将 Graphic 作为输入并且不返回任何内容(Void 或 ())。

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；它需要一个 Graphic 并且不返回任何内容。

因此 speakTool.onGraphicTouched = speakText(graphic: ) 将指向函数的指针分配给 onGraphicTouched，当图形被触摸时，speakTool 将调用 onGraphicTouched(someGraphic)，这将调用 speakText(graphic: someGraphic)。

您可以在 Apple's Swift Guide. 中关于函数类型 的部分阅读更多相关信息