swift - 带量程的开关盒

Question

我正在学习 Swift，并尝试在观看视频之前自己编写 Ryan Wenderlich 的游戏“Bullseye”。

我需要根据用户与目标数字的接近程度来给用户打分。我试着计算差异，然后检查范围并给用户分​​数，这就是我用 If-else 所做的(不能用 switch case 做):

private func calculateUserScore() -> Int {
    let diff = abs(randomNumber - Int(bullsEyeSlider.value))
    if diff == 0 {
        return PointsAward.bullseye.rawValue
    } else if diff < 10 {
        return PointsAward.almostBullseye.rawValue
    } else if diff < 30 {
        return PointsAward.close.rawValue
    }
    return 0 // User is not getting points. 
}

有没有办法更优雅地完成它或使用 Switch-Case？我不能只做 diff == 0 例如在 switch case 的情况下，因为 xCode 给我一条错误消息。

Answer 1

这应该有效。

private func calculateUserScore() -> Int {
    let diff = abs(randomNumber - Int(bullsEyeSlider.value))
    switch diff {
    case 0:
        return PointsAward.bullseye.rawValue
    case 1..<10:
        return PointsAward.almostBullseye.rawValue
    case 10..<30:
        return PointsAward.close.rawValue
    default:
        return 0
    }
}

它在 The Swift Programming Language 中在 Control Flow -> Interval Matching 下预订。