负数时 Swift 加倍奇怪

Question

Xcode 8.2.1、Swift 3。

import UIKit

extension Double {
    /// Linear interpolation.
    /// Converts a number in one range to its equivalent in another range.
    func interpolate(from: ClosedRange<Int>, to: ClosedRange<Int>) -> Double {
        let oldValue = self
        let offset = Double(to.lowerBound - from.lowerBound)
        let expansion = Double(to.upperBound - to.lowerBound) / Double(from.upperBound - from.lowerBound)

        print("\nvalue", oldValue)
        print("Sign", self.sign)
        print("offset", offset)
        print("expansion", expansion)
        print("oldLowerbound", Double(from.lowerBound))

        let newValue = (oldValue - Double(from.lowerBound)) * expansion + offset
        return newValue
    }
}
print( 1.0.interpolate(from: 0...10, to: 50...70)) // Prints 52.0.  Correct
print(-1.0.interpolate(from: 0...10, to: 50...70)) // Prints -52.0.  Expected 48.0.
print((-1.0).interpolate(from: 0...10, to: 50...70)) // Prints 48. Correct.

当单元测试在最(看似)微不足道的代码中揭示意想不到的问题时，它们会为自己付出代价:

(上面的代码将在 iOS playground 中运行。)

有没有一种方法可以扩展Double以使第二个打印语句按预期工作(无需像第三个打印语句那样使用括号)？

Answer 1

Martin R 是正确的，这是评估顺序的问题，遗憾的是(据我所知)中缀运算符(在本例中为 -)不能指定优先级。我想到的唯一方法是自定义插值运算符

infix operator ~|
func ~|(lhs: Double, rhs: (from: ClosedRange<Int>, to: ClosedRange<Int>)) -> Double {
  return lhs.interpolate(from: rhs.from, to: rhs.to)
}

所以结果是:

-1.0 ~| (from: 0...10, to: 50...70) // prints 48

并且为了防止 Double 扩展方法 interpolate 被直接调用，它可以被文件私有(private)地封装在同一个文件中。