我正在按照刚刚为 Swift 发布的 CS193p 讲座创建一个计算器应用程序。如何将 Pi 既作为操作数 & 又作为操作来实现？

Question

我正在努力完成讲座中发布的作业，但当然......我被困在第一个作业上，我讨厌在无法解决问题的情况下继续前进。

我在将 π 作为操作和和操作数时遇到问题。我让它作为一个操作工作（它所做的只是 return π 的值）。例如。 ' π Enter π Enter + ' 结果是 6.28... 但是如果我说 ' π Enter π 输入 + π 输入 ÷ ' 结果是 1.0 而应该是 2.0

我搜索了 Whosebug，除了 Objective-C 解决方案外，一无所获。感谢任何帮助。

这是当您点击操作时 ViewController 中触发的操作：

        @IBAction func operate(sender: UIButton) {
                let operand = sender.currentTitle!
                if (userIsInTheMiddleOfTypingANumber){
                    enter()
                }
                if let operation = sender.currentTitle{
                    if let result = brain.performOperation(operation){
                        displayValue = result
                    } else {
                        displayValue = 0
                    }
                }
            }

这是模型中的代码：

        private enum Op: Printable  {
    case Operand(Double)
    case UnaryOperator(String, Double -> Double)
    case BinaryOperator(String, (Double, Double) -> Double)
    case ConstantOperator (String, Double)

    var description: String {
        get {
            switch self {
            case .Operand(let operand):
                return "\(operand)"
            case .UnaryOperator(let symbol, _ ):
                return symbol
            case .BinaryOperator(let symbol, _ ):
                return symbol
            case .ConstantOperator(let symbol, _ ):
                return symbol
            }
        }
    }
}
func performOperation(symbol: String) -> Double?  {
    if let operation = knownOps[symbol]{
        opStack.append(operation)
    }
    return evaluate()
}

private var opStack = [Op]()
private var knownOps = [String:Op]()

init(){
    func learnOp(op: Op){
        knownOps[op.description] = op
    }

    learnOp(Op.BinaryOperator("✕", *))
    //knownOps["✕"] = Op.BinaryOperator("✕", *)
    knownOps["÷"] = Op.BinaryOperator("÷") { / [=12=]}
    knownOps["+"] = Op.BinaryOperator("+", +)
    knownOps["−"] = Op.BinaryOperator("−") { - [=12=]}
    knownOps["√"] = Op.UnaryOperator("√", sqrt)
    learnOp(Op.UnaryOperator("cos", cos))
    learnOp(Op.UnaryOperator("sin", sin))
    learnOp(Op.ConstantOperator("∏", M_PI))
}

private func evaluate(ops: [Op]) -> (result: Double?, remainingOps: [Op]){
    if !ops.isEmpty {

        var remainingOps = ops
        let op = remainingOps.removeLast() //get the first op off the stack

        switch op {
        case .Operand(let operand):
            return (operand, remainingOps)

        case .UnaryOperator(_, let operation):
            let operationEvaluation = evaluate(remainingOps)
            if let operand = operationEvaluation.result {
                return(operation(operand), operationEvaluation.remainingOps)
            }

        case .BinaryOperator(_ , let operation):
            let op1Eval = evaluate(remainingOps)
            if let op1 = op1Eval.result{
                let op2Eval = evaluate(op1Eval.remainingOps)
                if let op2 = op2Eval.result {
                    return (operation(op1, op2), op2Eval.remainingOps)
                }
            }

        case .ConstantOperator(_, let value):
            return(value, remainingOps)

        }
    }
    return(nil, ops)
}
func evaluate() -> Double? {
    let (result, remainder) = evaluate(opStack)
    println("\(opStack) = \(result) with \(remainder) left over.")
    return result
}

注意：鉴于教师希望我们遵循的结构，我正在尝试完成作业的必要任务。我在这里的尝试（尝试添加 ‖ 作为运算符）在某些情况下有效，但并非全部...

以下是作业中关于此任务的内容： "The value of π is available via the expression M_PI. E.g. let x = M_PI. You can think of π as an operand or you can think of it as an operation (i.e. a new kind of operation that takes no arguments off the stack but returns a value). Up to you. But, either way, it’d be nice to be able to add other constants to your Calculator with a minimum of code."

我尝试了操作数方法，但我打破了 MVC 范式，所以我停止了。

Answer 1

如果您按 π 并输入，那么您实际上是将 pi 放入 opStack 两次。所以你实际上是在做 π π π π + π π ÷。那有意义吗？为了将 pi 添加到自身，您需要推动 π π +。如果你想把 pi 加在一起然后除以 pi，那就是 π π + π ÷。

我正在按照刚刚为 Swift 发布的 CS193p 讲座创建一个计算器应用程序。如何将 Pi 既作为操作数 & 又作为操作来实现？

I'm creating a calculator app by following the CS193p lectures just released for Swift. How can I implement Pi as both an operand & and operation?

cs193p

swift

xcode6

ios8