学习迅速尝试理解如何使代码更简单

Question

是否可以使这个无效代码更简单

它基本上从2个盒子中获取2个值，具体取决于所选的 操作员进行计算。

   @IBAction func enterpressedinbox2(sender: AnyObject) {
    let value1toint = Int(value1.stringValue)
    let value2toint = Int(value2.stringValue)
    var result = 0;

    if operatorselection.stringValue == "+"{
        result = value1toint! + value2toint!
    }
    if operatorselection.stringValue == "-"{
        result = value1toint! - value2toint!
    }
    if operatorselection.stringValue == "*"{
        result = value1toint! * value2toint!
    }
    if operatorselection.stringValue == "/"{
        result = value1toint! / value2toint!
    }
    let resultinttostring = String(result)

    resultlabel.stringValue = ": " + resultinttostring
    view.window!.title = "The result is " + resultinttostring
}

Answer 1

我会用函数字典做这样的事情：

@IBAction func enterpressedinbox2(sender: AnyObject) {
    let ops : [String: (Int, Int)->Int] = [ "+" : { x, y in x + y }, 
                                            "-" : { x, y in x - y },
                                            "*" : { x, y in x * y }, 
                                            "/" : { x, y in x / y } ]
    if let x = Int(value1.stringValue),
       let y = Int(value2.stringValue),
       let f = ops[operatorselection.stringValue] {
           let result = String(f(x, y))
           resultlabel.stringValue = ": " + result
           view.window!.title = "The result is " + result
    }
}