Stanford CS193p iOS开发 课堂作业(1)

2015-12-01  本文已影响246人  Dominic1992

在进行了第一课Logistics,iOS8 Overview.和第二课More Xcode and Swift,MVC.两课的学习后有了第一次的课堂作业,要求完善计算器DEMO的功能.

基本要求

额外要求

2015年11月30日

实现了"sin","cos","π"的运算

    @IBAction func operate(sender: UIButton) { //创建运算符操作方法
        let operation = sender.currentTitle!  //识别运算符
        if userIsInTheMiddleOfTypingANumber { //可省略一次enter的点击操作,如不写这段代码则是"6  enter  6  enter  X"最后输出36,有这段则是"6  enter  6  X"
            enter()
        }
        switch operation {  //利用switch语句实现运算符的操作
        case "×" :performOperation {$0 * $1}  //利用闭包简化代码
        case "÷" :performOperation {$1 / $0}  //因为$1为倒数第二个数,$0为最后一个数,故除法和减法要用$1 /- $0  而乘法和加法则不必考虑顺序
        case "+" :performOperation {$0 + $1}
        case "−" :performOperation {$1 - $0}
        case "√" :performOperation2 {sqrt($0)}
        case "sin" :performOperation2 {sin($0)}
        case "cos" :performOperation2 {cos($0)}
        case "π" :performOperation3(M_PI)
        default: break
        }
    }

    func performOperation(operation: (Double,Double) -> Double) {
        if operandStack.count >= 2 {
            displayValue = operation(operandStack.removeLast(), operandStack.removeLast()) //获取了倒数第一个和倒数第二个数,并使其出栈
            enter()
        }
    }
    
    func performOperation2 (operation: Double -> Double) {
        if operandStack.count >= 1 {
            displayValue = operation(operandStack.removeLast())
            enter()
        }
    }
    
    func performOperation3(operation: Double) {
        displayValue = operation
        enter()
    }

2015年12月2日

实现了"C"清除按钮的功能,由于其他功能已暂时的学习程度不能完美解决,故先继续进行下一步的学习,之后再实现未完成的功能

    @IBAction func clearButton(sender: UIButton) {
        operandStack.removeAll()
        display.text = "0"
        userIsInTheMiddleOfTypingANumber = false
    }

2015年12月3日

实现"←"删除按钮的功能

    //实现删除按钮
    @IBAction func backSpace(sender: UIButton) {
        if display.text?.characters.count > 0 {
            display.text = String((display.text!).characters.dropLast())
        }
        if display.text?.characters.count == 0 {
            display.text = "0"
            userIsInTheMiddleOfTypingANumber = false
            
        }
    }

实现"+/−"正负按钮转换功能

if display.text!.hasPrefix("−") {
            display.text?.removeAtIndex(display.text!.startIndex)
        } else {
            display.text = "−" + display.text!
        }

3.实现栈中的正负号改变则需增加运算方法,在func operate()中先判断是否点击了"+/−"button,如点击,实现changeDisplaySign(),再在switch中加入case "+/−" :performOperation2 { -$0 }
代码如下:

    @IBAction func operate(sender: UIButton) { //创建运算符操作方法
        let operation = sender.currentTitle!  //识别运算符
        if userIsInTheMiddleOfTypingANumber { //可省略一次enter的点击操作,如不写这段代码则是"6  enter  6  enter  X"最后输出36,有这段则是"6  enter  6  X"
            if operation == "+/−" {
                changeDisplatSign()
                return
            }
            enter()
        }
        switch operation {  //利用switch语句实现运算符的操作
        case "×" :performOperation {$0 * $1}  //利用闭包简化代码
        case "÷" :performOperation {$1 / $0}  //因为$1为倒数第二个数,$0为最后一个数,故除法和减法要用$1 /- $0  而乘法和加法则不必考虑顺序
        case "+" :performOperation {$0 + $1}
        case "−" :performOperation {$1 - $0}
        case "√" :performOperation2 {sqrt($0)}
        case "sin" :performOperation2 {sin($0)}
        case "cos" :performOperation2 {cos($0)}
        case "π" :performOperation3(M_PI)
        case "+/−" :performOperation2 { -$0 }
        default: break
        }
    }
    func performOperation2 (operation: Double -> Double) {
        if operandStack.count >= 1 {
            displayValue = operation(operandStack.removeLast())
            enter()
        }
    }

    //实现正负转换按钮功能
    func changeDisplatSign() {
        if display.text!.hasPrefix("−") {
            display.text?.removeAtIndex(display.text!.startIndex)
        } else {
            display.text = "−" + display.text!
        }
    }
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