关于swift:Swift-Learning-Summary-Opaque-Type

Opaque Type

A function with an opaque type hides its return value’s type information. Hiding type information at some boundaries between a module and code that calls into the module. Unlike returning a value whose type is a protocol type, opaque type preserve type identity —the compile has access to the type information, but clients of the module don’t.

The Situation

Here we have a Shape protocol.

protocol Shape {func draw() -> String
}

The struct Triangle conform to the Shape. Describe how to draw().

struct Triangle: Shape {
    var size: Int
    func draw() -> String {var result: [String] = []
        for length in 1...size {result.append(String(repeating: "*", count: length))
        }
        return result.joined(separator: "\n")
    }
}

let smallTriangle = Triangle(size: 4)
print(smallTriangle.draw())
// Print:
// *
// **
// ***
// ****

The struct FlippedShape conform to the Shape and it need an injection in a type of Shape.

struct FlippedShape<T: Shape>: Shape{
    var shape: T
    func draw() -> String {let lines = shape.draw().split(separator: "\n")
        return lines.reversed().joined(separator: "\n")
    }
}

let filppingShape = FlippedShape<Triangle>(shape: smallTriangle)
print(filppingShape.draw())
// Print:
// ****
// ***
// **
// *

The JoinedShape conform to the Shape and it need two injection in type of Shape. It use the generic type T and U.

struct JoinedShape<T: Shape, U: Shape>: Shape{
    var top: T
    var bottom: U
    func draw() -> String {return top.draw() + "\n" + bottom.draw()}
}
let joinedShape = JoinedShape(top: smallTriangle, bottom: filppingShape)
print(joinedShape.draw())
// Print:
// *
// **
// ***
// ****
// ****
// ***
// **
// *

Returning an Opaque Type

The opaque type like being the reverse of a generic type.

In generic, the function return a type that depends on its caller:

func max<T>(_ x: T, _ y: T) -> where T: Comparable {...}

Use opaque type. It return a some type and don’t exposing the underlying type of that shape. It only focus on the return type, not the specific type.

struct Square: Shape {
    var size: Int
    func draw() -> String {let line = String(repeating: "*", count: size)
        let result = Array<String>(repeating: line, count: size) count: <#T##Int#>)
        return  result.joined(separator: "\n")
    }
}

func makeTrapezoid() -> some Shape {let top = Triangle(size: 2)
    let middle = Square(size: 2)
    let bottom = FlippedShape(shape: top)
    let trapezoid = JoinedShape(top: top, bottom: JoinedShape(top: middle, bottom: bottom))
    return trapezoid
}

// Here we get a trapezoid, it's something that conform to the Shape and we can only use it as shape, the client  can't access the underlying information of this shape.
let trapezoid = makeTrapezoid()
print(trapezoid.draw())

// Print:
// *
// **
// **
// **
// **
// *

Combine Opaque Return Type with Generics.

func flip<T: Shape>(_ shape: T) -> some Shape {return FlippedShape(shape: shape)
}
func join<T: Shape, U: Shape>(_ top: T, _ bottom: U) -> some Shape {return JoinedShape(top: top, bottom: bottom)
}
let opaqueJoinedTriangle = join(smallTriangle, flip(smallTriangle))
print(opaqueJoinedTriangle.draw())

// Print:
// *
// **
// ***
// ****
// ****
// ***
// **
// *

All the possible opaque return in a function must have the same type.

// Here is an example in error 

func invalidFlip<T: Shape>(_ shape: T) -> some Shape {
    if shape is Square {return shape}
    return FlippedShape(shape: shape)
}

One way to avoid return different type is to move this Square case into the FlippedShape implementation.

struct FlippedShape<T: Shape>: Shape{
    var shape: T
    func draw() -> String {
        if shape is Square {return shape.draw()
        }
        
        let lines = shape.draw().split(separator: "\n")
        return lines.reversed().joined(separator: "\n")
    }
}

Using generics in an opaque return type.

func repeatObj<T: Shape>(shape: T, count: Int) -> some Collection {return Array<T>(repeating: shape, count: count)
}

Differences Between Opaque Types and Protocol Types

Using protocol type

It can return different type that conform to Shape , it makes a much looser API than opaque return type make.

// Protocol type
func protoFlip<T: Shape>(_ shape: T) -> Shape {
    if shape is Square {return shape}

    return FlippedShape(shape: shape)
}

The less specific return type information means that the operation that depends on type information aren’t available on the return value.

let protoFlipTriangle = protoFlip(smallTriangle)
let sameThing = protoFlip(smallTriangle)
print(protoFlipTriangle == sameThing)   // Error they are 'Shape', 'Shape' has no func to check if they are equal, operator '==' cannot be applied to two 'Shape'  

The opaque types preserve the identity of the underlying type.

protocol Container {
    associatedtype Item 
    var count: Int {get}
    subscript(i: Int) -> Item {get}
}

extension Array: Container {}

Here we:

• can’t use Container as the return type of a function. Because the protocol has an associated type.
• And can’t use it as constraint in a generic return type. Because there isn’t enough information outside the function body to infer what the generic type needs to be.

// Error: Protocol with associated types can't be used as a return type.
func makeProtocolContainer<T>(item: T) -> Container {return [item]
}

// Error: Not enough information to infer C, it has associate type
func makeProtocolContainer<T, C: Container>(item: T) -> C {return [item]
}

Using the opaque type some Container as a return type. It means that the function return a container, but declines to specify the container’s type.

func makeOpaqueContainer<T>(item: T) -> some Container {return  [item]
}

let opaqueContainer = makeOpaqueContainer(item: 12)
let twelve = opaqueContainer[0]
print(type(of: twelve))  // Int