Swift 如何防止某些泛型重载，例如整数类型的 Range.subscript？

Question

Swift 类型 Range 的文档指出：

Like other collections, a range containing one element has an endIndex that is the successor of its startIndex; and an empty range has startIndex == endIndex.

Axiom: for any Range r, r[i] == i.

Therefore, if T has a maximal value, it can serve as an endIndex, but can never be contained in a Range<T>.

It also follows from the axiom above that (-99..<100)[0] == 0. To prevent confusion (because some expect the result to be -99), in a context where T is known to be an integer type, subscripting with T is a compile-time error::
// error: could not find an overload for 'subscript'…   
println( Range<Int>(start:-99, end:100)[0] )
However, subscripting that range still works in a generic context::
func brackets<T:ForwardIndexType>(x: Range<T>, i: T) -> T {
  return x[i] // Just forward to subscript
}
println(brackets(Range<Int>(start:-99, end:100), 0)) // prints 0

我们如何在我们自己的泛型中实现这种行为类？仅从 subscript:

的定义我无法理解它是如何工作的

subscript (position: T) -> T { get }
subscript (_: T._DisabledRangeIndex) -> T { get }

谁能给我解释一下？

Answer 1

以下代码的行为与内置 Range 相同。

struct MyRange<T: ForwardIndexType> {
    subscript (position: T) -> T {
        return position
    }
    subscript (_: T._DisabledRangeIndex) -> T {
        fatalError()
    }
}

MyRange<String.Index>()["foo".startIndex] // no problem
MyRange<Int>()[0] // < [!] error: could not find an overload for 'subscript' that accepts the supplied arguments

好像Int._DisabledRangeIndex是Int，Int16._DisabledRangeIndex是Int16等等

当 T._DisabledRangeIndex 与 T 相同时，MyRange<T> 最终有 2 个冲突的 subscript 定义导致编译错误。

Swift 如何防止某些泛型重载，例如整数类型的 Range.subscript？

How does Swift prevent certain generic overloads, like Range.subscript for integer types?

generics

overloading

swift