Swift 中的计算器

Question

我对这个计算器有一些逻辑问题。当我在第一次执行后按下“=”时，它将使用相同的操作数执行两次。

例如：

"1"+"2"="3"

“3”+“2”，然后在我按下另一个操作数之前输出->“5”。

这里是源代码：

https://drive.google.com/file/d/0B1a-AefbM9rOUTdyVmdJOWhNNnc/view?usp=sharing

import UIKit

class ViewController: UIViewController {

    @IBOutlet weak var display: UILabel!
    
    var MiddleTyping = false
    
    @IBAction func appandDigit(sender: UIButton) {
        let digit = sender.currentTitle!
        if MiddleTyping{
            display.text = display.text! + digit
        }else{
            display.text = digit
            MiddleTyping = true
        }
    }
    
    
    
    var operandStack = [Double]()
    var binaryoperation = [String]();
    var firstTimeInputNumber = true
    
    var displayValue: Double{
        get{
            return NSNumberFormatter().numberFromString(display.text!)!.doubleValue
        }
        set{
            display.text = "\(newValue)"
            MiddleTyping = false
        }
    }
    @IBAction func Operate(sender: UIButton) {
        MiddleTyping = false
        var equalOperand = ""
        var Operation = ""
        if sender.currentTitle == "="{
             equalOperand = binaryoperation.last ?? ""
        }else{
              Operation = sender.currentTitle!
        }
        if operandStack.count == 1 && binaryoperation.count != 0{
            operandStack.append(displayValue)
        }
        if operandStack.count == 2{
            if equalOperand == binaryoperation.last ?? "" {
                switch equalOperand{
                case "×":
                    displayValue = operandStack[0] * operandStack[1]
                    operandStack[0] = displayValue
                case "÷":
                    displayValue = operandStack[0] / operandStack[1]
                    operandStack[0] = displayValue
                case "+":
                    displayValue = operandStack[0] + operandStack[1]
                    operandStack[0] = displayValue
                case "−":
                    displayValue = operandStack[0] - operandStack[1]
                    operandStack[0] = displayValue
                default: break
                }
            }else{
                let Operation = binaryoperation.last ?? "";
                switch Operation {
                case "×":
                    displayValue = operandStack[0] * operandStack[1]
                    operandStack[0] = displayValue
                    operandStack.removeLast()
                case "÷":
                    displayValue = operandStack[0] / operandStack[1]
                    operandStack[0] = displayValue
                    operandStack.removeLast()
                case "+":
                    displayValue = operandStack[0] + operandStack[1]
                    operandStack[0] = displayValue
                    operandStack.removeLast()
                case "−":
                    displayValue = operandStack[0] - operandStack[1]
                    operandStack[0] = displayValue
                    operandStack.removeLast()
                default: break
                }
            }
        }
        if displayValue != 0 {
            if firstTimeInputNumber{
                operandStack.append(displayValue)
            }
            firstTimeInputNumber = false
        }
        if Operation != ""{
            binaryoperation.append(Operation)
        }
        println("push op \(operandStack)")
        println("push op \(binaryoperation)")
    }
    @IBAction func ClearButton() {
        operandStack.removeAll()
        binaryoperation.removeAll()
        display.text = "0"
        firstTimeInputNumber = true
        MiddleTyping = false
        
    }
}

Answer 1

我所能想象的只是条件binaryoperation.count != 0 可能是 binaryoperation.count == 1，也就是你在计算的时候不要去掉操作符