Update: The answer below still applies in principle, but Swift 4 completed a redesign of the numeric protocols, such that adding your own is often unnecessary. Take a look at the standard library's numeric protocols before you build your own system.
This actually isn't possible out of the box in Swift. To do this you'll need to create a new protocol, declared with whatever methods and operators you're going to use inside your generic function. This process will work for you, but the exact details will depend a little on what your generic function does. Here's how you'd do it for a function that gets a number n
and returns (n - 1)^2
.
First, define your protocol, with the operators and an initializer that takes an Int
(that's so we can subtract one).
protocol NumericType {
func +(lhs: Self, rhs: Self) -> Self
func -(lhs: Self, rhs: Self) -> Self
func *(lhs: Self, rhs: Self) -> Self
func /(lhs: Self, rhs: Self) -> Self
func %(lhs: Self, rhs: Self) -> Self
init(_ v: Int)
}
All of the numeric types already implement these, but at this point the compiler doesn't know that they conform to the new NumericType
protocol. You have to make this explicit -- Apple calls this "declaring protocol adoption with an extension." We'll do this for Double
, Float
, and all the integer types:
extension Double : NumericType { }
extension Float : NumericType { }
extension Int : NumericType { }
extension Int8 : NumericType { }
extension Int16 : NumericType { }
extension Int32 : NumericType { }
extension Int64 : NumericType { }
extension UInt : NumericType { }
extension UInt8 : NumericType { }
extension UInt16 : NumericType { }
extension UInt32 : NumericType { }
extension UInt64 : NumericType { }
Now we can write our actual function, using the NumericType
protocol as a generic constraint.
func minusOneSquared<T : NumericType> (number : T) -> T {
let minusOne = number - T(1)
return minusOne * minusOne
}
minusOneSquared(5) // 16
minusOneSquared(2.3) // 1.69
minusOneSquared(2 as UInt64) // 1
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