正確應用封裝

Question

我一直在研究封裝的概念，並發現了一些關於像this one和this one這樣的話題的體面線索。但是我一直無法找到我所具有的特定問題的答案。我將在Swift中使用一個示例。

假設你有一個對象，它的類型是旅遊路徑的：

class RoadTrip() { 
    private var duration: Double 
    private var totalMiles: Double 
    private var gallonsOfFuel: Double 

    var averageMilesPerGallon: Double 
} 

現在讓我們假設該應用程序會計算averageMilesPerGallon這是唯一的公共財產：

func calculateAverageMilePerGallon() -> Double { 
    let mpg = totalMiles/gallonsOfFuel 

    return mpg 
} 

如若計算每加侖平均英里數是RoadTrip對象的私有方法，該方法執行並更新其averageMilesPerGallon，或者在單獨的工具類中通過另一方法執行計算，然後更新averageMilesPerGallon RoadTrip對象的屬性是否使用了會設置值的增變量方法？

編輯：這是我的單個類，其中包含我的應用程序的基本計算。我走近它基於我在iTunes上斯坦福大學的過程中瞭解到這樣一來，但是我開始覺得我的情況下，我應該搬到這在很大程度上給我LiftEvent類：

infix operator ^^ { } 
func ^^ (radix: Double, power: Double) -> Double { 
    return Double(pow(Double(radix), Double(power))) 
} 

class CalculatorBrain: NSObject { 
    var weightLifted: Double? 
    var repetitions: Double? 
    var oneRepMax: Double? 
    let dataManager = LiftEventDataManager() 

    func calculateOneRepMax(weightLifted: Double, repetitions: Int) -> Double { 
     let preferredFormulaID = UserDefaultsManager.sharedInstance.preferredFormula! 
     let formulas = dataManager.fetchSelectableFormulas() 
     let formulaName = formulas[preferredFormulaID].formulaName 

     switch formulaName { 
     case "Epley": 
      oneRepMax = weightLifted * (1 + Double(repetitions)/30.0) 
      return oneRepMax! 
     case "Baechle": 
      oneRepMax = weightLifted * (36/(37 - Double(repetitions))) 
      return oneRepMax! 
     case "Brzychi": 
      oneRepMax = weightLifted * (1 + (0.033 * Double(repetitions))) 
      return oneRepMax! 
     case "Lander": 
      oneRepMax = 100 * weightLifted/(101.3 - (2.67123 * Double(repetitions))) 
      return oneRepMax! 
     case "Lombardi": 
      oneRepMax = weightLifted * (Double(repetitions) ^^ 0.10) 
      return oneRepMax! 
     case "Mayhew": 
      oneRepMax = 100 * weightLifted/(52.2 + (41.9 * (2.71828 ^^ (-0.055 * Double(repetitions))))) 
      return oneRepMax! 
     case "O'Conner": 
      oneRepMax = weightLifted * (1 + 0.025 * Double(repetitions)) 
      return oneRepMax! 
     default: 
      return 0.0 
     } 
    } 


    private func calculatePercentOfWeight(maxWeight: Double, percent: Double) -> Double { 
     return maxWeight * percent 
    } 

    func calculateWeightPercentages(maxWeight: String) -> [Int: Double] { 
     let weightPercentages = [1.0, 0.95, 0.90, 0.85, 0.80, 0.75, 0.70, 0.65, 0.60, 0.55, 0.50, 0.45, 0.40, 0.35, 0.30, 0.25] 

     var percentages = [Int: Double]() 
     for percent in weightPercentages { 
      let integerPercent = Int(percent * 100) 
      percentages[integerPercent] = calculatePercentOfWeight(Double(maxWeight)!, percent: percent) 
     } 
     return percentages 
    } 

    func convertBetweenUnits(fromUnit: Int, toUnit: Int, value: Double) -> Double { 
     let units = dataManager.fetchUnits() 
     let from = units[fromUnit].conversionRatio as Double 
     let to = units[toUnit].conversionRatio as Double 
     let result = Double(value * to/from) 
     return result 
    } 
} 

Answer 1

我認爲這是理想的用例的計算性能：

class RoadTrip { 
    private let duration: Double 
    private let totalMiles: Double 
    private let gallonsOfFuel: Double 

    private var averageMilesPerGallon: Double { 
     return totalMiles/gallonsOfFuel 
    } 
}