可以在字符串的插值中包括一个方法吗？ （迅速）

Question

我试图将我的commercial()类的Cars方法包括在我的finalNegotiation()类的PriceNegotiation方法执行的字符串插值中。那可能吗？ 我尝试使用.commercial()，也尝试了使用super.commerical()

的代码

class Cars  {
    var make = ""
    var model = ""
    var year = 0

    init(carMake make:String, carModel model:String, carYear year:Int) {
        self.make = make
        self.model = model
        self.year = year
    }
    func commercial() {
        print("This car is a \(year) \(make) \(model)")
    }

}




class PriceNegotation: Cars {
    var price:Double = 0


    init(desiredBuyerPrice price:Double,carMake make:String, carModel model:String, carYear year:Int ) {
        self.price = price

    super.init(carMake: make, carModel: make, carYear: year)


    }

    func finalNegotiation() {
        let dealerPrice = price * 1.5
        print("Since \(super.commercial()) the asking price is \(dealerPrice)")
    }

}

Answer 1

调用commercial()时，它会打印文本并退出该功能。为了实现您的目标，请使commercial()函数包含一个返回值。这是一个例子。

class Cars  {
    var make = ""
    var model = ""
    var year = 0

    init(carMake make:String, carModel model:String, carYear year:Int) {
        self.make = make
        self.model = model
        self.year = year
    }

    func commercial()->String {
       return "This car is a \(year) \(make) \(model)"
    }
}
class PriceNegotation: Cars {
    var price:Double = 0

    init(desiredBuyerPrice price:Double,carMake make:String, carModel model:String, carYear year:Int ) {
        self.price = price

        super.init(carMake: make, carModel: make, carYear: year)
    }

    func finalNegotiation() {
        let dealerPrice = price * 1.5
        let commercialOutput = commercial()
        print("Since \(commercialOutput) the asking price is \(dealerPrice)")
    }
}

这是通过插入将commercial()函数String的输出放在另一个字符串中的方法。以前该函数未返回任何内容，因此看起来好像该函数不起作用。另一方面，这应该起作用。让我知道是否可以。