ios - 如何扩展算术运算符以支持可选值 - Swift

Question

在下面的示例中，

let value1: Int? = 23
let value2: Int = 20

let answer = value1 + value2 // Compiler warning that + operator cannot be applied to Int? and Int

所以我必须将代码更改为

if let value1 = value1 {
    let answer = value1 + value2 
}

如何为 + 创建一个也支持可选值的扩展？在这种情况下，它应该给出 nil 作为输出。

如果运算有多个操作数怎么办？

let value1: Int? = 2

let answer = value1 + 3.0

Answer 1

实际上，您只需要找到正确的协议(protocol)类型来约束泛型类型即可。之后的实现就很简单了:

// plus and minus is supported by AdditiveArithmetic
func +<T: AdditiveArithmetic>(lhs: T?, rhs: T?) -> T? {
    return lhs.flatMap { x in rhs.map { y in x + y } }
    /* the above is just a more "functional" way of writing 
    if let x = lhs, let y = rhs {
        return x + y
    } else {
        return nil
    }
    */
}

func -<T: AdditiveArithmetic>(lhs: T?, rhs: T?) -> T? {
    return lhs.flatMap { x in rhs.map { y in x - y } }
}

// times is supported by Numeric
func *<T: Numeric>(lhs: T?, rhs: T?) -> T? {
    return lhs.flatMap { x in rhs.map { y in x * y } }
}

// divide is not supported by a single protocol AFAIK
func /<T: BinaryInteger>(lhs: T?, rhs: T?) -> T? {
    return lhs.flatMap { x in rhs.map { y in x / y } }
}

func /<T: FloatingPoint>(lhs: T?, rhs: T?) -> T? {
    return lhs.flatMap { x in rhs.map { y in x / y } }
}

要使 value1 + 3.0 工作，您必须执行以下操作:

func +<T: BinaryInteger, U: FloatingPoint>(lhs: T?, rhs: U?) -> U? {
    return lhs.flatMap { x in rhs.map { y in U(x) + y } }
}

但违反现有的限制通常不是一个好主意。我不推荐这样做。