Sco*_*ser 0 enums enumeration ios swift swift2
我不确定这是否可行,但我想知道是否可以将嵌套的枚举值转换为单个变量,以便将它们传递给函数参数.例:
enum Food {
enum Vegetables: String {
case Spinach = "Spinach"
case GreenBeans = "Green Beans"
}
enum Fruit: String {
case Apples = "Apples"
case Grapes = "Grapes"
}
}
func eat(meal:Food.Vegetables) {
print("I just ate some \(meal.rawValue)")
}
func eat(meal:Food.Fruit) {
print("I just ate some \(meal.rawValue)")
}
eat(Food.Fruit.Apples) //"I just ate some Spinach\n"
eat(Food.Vegetables.Spinach) //"I just ate some Apples\n"
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这里的所有东西都应该可行,但我正在尝试将我的两个吃功能合并为1.有没有办法做到这一点?我想它会涉及一个变量,它代表我可以传递给一个eat函数的所有嵌套变量类型.就像是:
func eat(fruitOrVegetable: Food.allNestedEnumeratorTypes) {
print("I just ate some \(fruitOrVegetable.rawValue)")
}
eat(Food.Vegetables.GreenBeans) //"I just ate some Green Beans\n"
eat(Food.Vegetables.Grapes) //"I just ate some Grapes\n"
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这可能吗?
你可以使用协议
protocol Vegetarian {
var rawValue : String { get }
}
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并将其添加到两个枚举
enum Vegetables: String, Vegetarian { ... }
enum Fruit: String, Vegetarian { ... }
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然后你可以申报 eat
func eat(meal:Vegetarian) {
print("I just ate some \(meal.rawValue)")
}
eat(Food.Vegetables.GreenBeans) //"I just ate some Green Beans\n"
eat(Food.Fruit.Grapes) //"I just ate some Grapes\n"
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