Swift将Integer转换为Int

Question

I am doing generics programming and have something that conforms to Integer . Somehow I need to get that into a concrete Int that I can use.

extension CountableRange
{
    // Extend each bound away from midpoint by `factor`, a portion of the distance from begin to end
    func extended(factor: CGFloat) -> CountableRange<Bound> {
        let theCount = Int(count) // or lowerBound.distance(to: upperBound)
        let amountToMove = Int(CGFloat(theCount) * factor)
        return lowerBound - amountToMove ..< upperBound + amountToMove
    }
}

The error here is on let theCount = Int(count) . Which states:

Cannot invoke initializer for type 'Int' with an argument list of type '(Bound.Stride)'

First the error could be more helpful because CountableRange defines its Bound.Stride as SignedInteger ( source ). So the error could have told me that.

So I know it is an Integer , but how do I actually make use of the Integer value?

Answer 1

You can use numericCast() to convert between different integer types. As the documentation states:

Typically used to do conversion to any contextually-deduced integer type.

In your case:

extension CountableRange where Bound: Strideable {

    // Extend each bound away from midpoint by `factor`, a portion of the distance from begin to end
    func extended(factor: CGFloat) -> CountableRange<Bound> {
        let theCount: Int = numericCast(count)
        let amountToMove: Bound.Stride = numericCast(Int(CGFloat(theCount) * factor))
        return lowerBound - amountToMove ..< upperBound + amountToMove
    }
}

The restriction Bound: Strideable is necessary to make the arithmetic lowerBound - amountToMove and upperBound + amountToMove compile.

Answer 2

这应该适用于从swift 3.0开始

let theCount:Int32 = Int32(count);

Answer 3

If you really need that Int , try this instead:

let theCount = Int(count.toIntMax())

The toIntMax() method return this integer using Swift's widest native signed integer type (ie, Int64 on 64-bits platforms).