小于或大于Swift switch语句

Question

我熟悉Swift中的switch语句，但想知道如何用switch替换这段代码：

if someVar < 0 {
    // do something
} else if someVar == 0 {
    // do something else
} else if someVar > 0 {
    // etc
}

Answer 1

这是一种方法。假设someVar是Int或其他Comparable，您可以选择将操作数分配给新变量。这允许您根据需要使用where关键字对其进行范围调整：

var someVar = 3

switch someVar {
case let x where x < 0:
    print("x is \(x)")
case let x where x == 0:
    print("x is \(x)")
case let x where x > 0:
    print("x is \(x)")
default:
    print("this is impossible")
}

这可以简化一下：

switch someVar {
case _ where someVar < 0:
    print("someVar is \(someVar)")
case 0:
    print("someVar is 0")
case _ where someVar > 0:
    print("someVar is \(someVar)")
default:
    print("this is impossible")
}

您还可以使用范围匹配完全避免使用where关键字：

switch someVar {
case Int.min..<0:
    print("someVar is \(someVar)")
case 0:
    print("someVar is 0")
default:
    print("someVar is \(someVar)")
}

Answer 2

使用Swift 5，您可以选择以下开关之一来替换if语句。

---

#1 Using switch with PartialRangeFrom and PartialRangeUpTo

let value = 1

switch value {
case 1...:
    print("greater than zero")
case 0:
    print("zero")
case ..<0:
    print("less than zero")
default:
    fatalError()
}

---

#2 Using switch with ClosedRange and Range

let value = 1

switch value {
case 1 ... Int.max:
    print("greater than zero")
case Int.min ..< 0:
    print("less than zero")
case 0:
    print("zero")
default:
    fatalError()
}

---

#3 Using switch with where clause

let value = 1

switch value {
case let val where val > 0:
    print("\(val) is greater than zero")
case let val where val == 0:
    print("\(val) is zero")
case let val where val < 0:
    print("\(val) is less than zero")
default:
    fatalError()
}

---

#4 Using switch with where clause and assignment to _

let value = 1

switch value {
case _ where value > 0:
    print("greater than zero")
case _ where value == 0:
    print("zero")
case _ where value < 0:
    print("less than zero")
default:
    fatalError()
}

---

#5 Using switch with RangeExpression protocol's ~=(_:_:) operator

let value = 1

switch true {
case 1... ~= value:
    print("greater than zero")
case ..<0 ~= value:
    print("less than zero")
default:
    print("zero")
}

---

#6 Using switch with Equatable protocol's ~=(_:_:) operator

let value = 1

switch true {
case value > 0:
    print("greater than zero")
case value < 0:
    print("less than zero")
case 0 ~= value:
    print("zero")
default:
    fatalError()
}

---

#7 Using switch with PartialRangeFrom, PartialRangeUpTo and RangeExpression's contains(_:) method

let value = 1

switch true {
case (1...).contains(value):
    print("greater than zero")
case (..<0).contains(value):
    print("less than zero")
default:
    print("zero")
}

Answer 3

引子盖下的switch语句使用~=运算符。所以这：

let x = 2

switch x {
case 1: print(1)
case 2: print(2)
case 3..<5: print(3..<5)
default: break
}

Desugars to this：

if 1          ~= x { print(1) }
else if 2     ~= x { print(2) }
else if 3..<5 ~= x { print(3..<5) }
else {  }

If you look at the standard library reference, it can tell you exactly what the ~= is overloaded to do：包括范围匹配，等同于等同的东西。 （不包括枚举大小写匹配，这是一种语言特性，而不是std lib中的函数）

您会看到它与左侧的直布尔值不匹配。对于那些比较，您需要添加where语句。

除非......你自己超载了~=操作员。 （通常不建议这样做）一种可能性是这样的：

func ~= <T> (lhs: T -> Bool, rhs: T) -> Bool {
  return lhs(rhs)
}

因此匹配一个函数，该函数在左侧返回一个布尔值，在右侧返回其参数。这是你可以用它来做的事情：

func isEven(n: Int) -> Bool { return n % 2 == 0 }

switch 2 {
case isEven: print("Even!")
default:     print("Odd!")
}

对于您的情况，您可能有一个如下所示的语句：

switch someVar {
case isNegative: ...
case 0: ...
case isPositive: ...
}

但现在你必须定义新的isNegative和isPositive函数。除非你超载一些运营商......

您可以将正常的中缀运算符重载为curried前缀或后缀运算符。这是一个例子：

postfix operator < {}

postfix func < <T : Comparable>(lhs: T)(_ rhs: T) -> Bool {
  return lhs < rhs
}

这将是这样的：

let isGreaterThanFive = 5<

isGreaterThanFive(6) // true
isGreaterThanFive(5) // false

将它与早期的函数结合使用，你的switch语句可能如下所示：

switch someVar {
case 0< : print("Bigger than 0")
case 0  : print("0")
default : print("Less than 0")
}

现在，你可能不应该在实践中使用这种东西：它有点狡猾。你（可能）更好地坚持使用where声明。也就是说，switch语句模式

switch x {
case negative:
case 0:
case positive:
}

要么

switch x {
case lessThan(someNumber):
case someNumber:
case greaterThan(someNumber):
}

似乎很常见，值得考虑。

Answer 4

您可以：

switch true {
case someVar < 0:
    print("less than zero")
case someVar == 0:
    print("eq 0")
default:
    print("otherwise")
}

Answer 5

由于有人已经发布case let x where x < 0:这里是someVar是Int的替代品。

switch someVar{
case Int.min...0: // do something
case 0: // do something
default: // do something
}

以下是someVar是Double的替代方案：

case -(Double.infinity)...0: // do something
// etc

Answer 6

这是范围的样子

switch average {
    case 0..<40: //greater or equal than 0 and less than 40
        return "T"
    case 40..<55: //greater or equal than 40 and less than 55
        return "D"
    case 55..<70: //greater or equal than 55 and less than 70
        return "P"
    case 70..<80: //greater or equal than 70 and less than 80
        return "A"
    case 80..<90: //greater or equal than 80 and less than 90
        return "E"
    case 90...100: //greater or equal than 90 and less or equal than 100
        return "O"
    default:
        return "Z"
    }

Answer 7

<0表达式不起作用（不再？）所以我最终得到了这个：

Swift 3.0：

switch someVar {
    case 0:
        // it's zero
    case 0 ..< .greatestFiniteMagnitude:
        // it's greater than zero
    default:
        // it's less than zero
    }

Answer 8

很高兴Swift 4解决了这个问题：作为3中的解决方法，我做了：

switch translation.x  {
    case  0..<200:
        print(translation.x, slideLimit)
    case  -200..<0:
        print(translation.x, slideLimit)
    default:
        break
    }

工作但不理想