如何简化冗长的Swift switch语句？六个同类代码块需精简

你这段代码是典型的D&D类游戏属性值转修正值的逻辑，功能没问题但冗余度极高，完全可以用数学公式+边界判断替代冗长的switch，还能轻松复用在六个属性上：

原始臃肿代码 #

let strStatResult = strStat
strengthModLabel.text = String(strStatResult)

switch strStatResult {
case let strStatResult where strStatResult == 1:
    strengthModLabel.text = "-5"
case let strStatResult where strStatResult == 2 || strStatResult == 3:
    strengthModLabel.text = "-4"
case let strStatResult where strStatResult == 4 || strStatResult == 5:
    strengthModLabel.text = "-3"
case let strStatResult where strStatResult == 6 || strStatResult == 7:
    strengthModLabel.text = "-2"
case let strStatResult where strStatResult == 8 || strStatResult == 9:
    strengthModLabel.text = "-1"
case let strStatResult where strStatResult == 10 || strStatResult == 11:
    strengthModLabel.text = "0"
case let strStatResult where strStatResult == 12 || strStatResult == 13:
    strengthModLabel.text = "+1"
case let strStatResult where strStatResult == 14 || strStatResult == 15:
    strengthModLabel.text = "+2"
case let strStatResult where strStatResult == 16 || strStatResult == 17:
    strengthModLabel.text = "+3"
case let strStatResult where strStatResult == 18 || strStatResult == 19:
    strengthModLabel.text = "+4"
case let strStatResult where strStatResult == 20 || strStatResult == 21:
    strengthModLabel.text = "+5"
case let strStatResult where strStatResult == 22 || strStatResult == 23:
    strengthModLabel.text = "+6"
case let strStatResult where strStatResult == 24 || strStatResult == 25:
    strengthModLabel.text = "+7"
case let strStatResult where strStatResult == 26 || strStatResult == 27:
    strengthModLabel.text = "+8"
case let strStatResult where strStatResult == 28 || strStatResult == 29:
    strengthModLabel.text = "+9"
case let strStatResult where strStatResult == 30:
    strengthModLabel.text = "+10"
default:
    strengthModLabel.text = "Error"
}

精简方案 #

先梳理规律：属性值和修正值的对应关系完全符合D&D官方公式：修正值 = floor((属性值 - 10) / 2)，利用这个公式可以彻底干掉所有case判断，还能封装成通用函数处理六个属性：

通用函数实现

func updateAbilityModLabel(_ label: UILabel, for abilityScore: Int) {
    // 判断属性值是否在合法范围（1-30）
    guard (1...30).contains(abilityScore) else {
        label.text = "Error"
        return
    }
    // 计算修正值
    let mod = Int(floor(Double(abilityScore - 10) / 2))
    // 格式化显示文本：0直接显示0，正数加+，负数正常显示
    label.text = mod == 0 ? "0" : mod > 0 ? "+\(mod)" : "\(mod)"
}

调用方式

六个属性只需要调用六次函数即可，再也不用写重复的switch：

// 力量修正
updateAbilityModLabel(strengthModLabel, for: strStat)
// 敏捷修正
updateAbilityModLabel(dexterityModLabel, for: dexStat)
// 体质修正
updateAbilityModLabel(constitutionModLabel, for: conStat)
// 智力修正
updateAbilityModLabel(intelligenceModLabel, for: intStat)
// 感知修正
updateAbilityModLabel(wisdomModLabel, for: wisStat)
// 魅力修正
updateAbilityModLabel(charismaModLabel, for: chaStat)

为什么不用循环？ #

循环在这里完全没必要——因为对应关系是明确的数学公式，直接计算比遍历判断效率高得多，而且代码更简洁易维护。如果后续属性范围调整，只需要修改公式或合法范围即可，不用改动一堆case。

内容的提问来源于stack exchange，提问作者Lucas Rider