TypeScript 条件返回类型 : works for one condition but not for two

Question

我有一个基于 discriminated union 创建形状的工厂函数形状参数，例如:

interface CircleArgs { type: "circle", radius: number };
interface SquareArgs { type: "square", length: number };

type ShapeArgs = CircleArgs | SquareArgs;

class Circle { constructor(_: CircleArgs) {}}
class Square { constructor(_: SquareArgs) {}}

type Shape = Circle | Square;

type ShapeOfArgs<Args extends ShapeArgs> = 
   Args extends CircleArgs? Circle :
   Square;

function createShape<Args extends ShapeArgs>(args: Args): ShapeOfArgs<Args> {
    switch (args.type) {
        case "circle": return new Circle(args);
        case "square": return new Square(args);
    }
}

这样做不仅有助于 TS 从参数中推断出正确的返回类型:

const circle1 = createShape({type: "circle", radius: 1}); // inferred as Circle
const square1 = createShape({type: "square", length: 1}); // inferred as Square

但如果我需要将调用包装在另一个函数中，也会正确传播返回类型:

class Container { insert(_: Shape): void {}; }

function createShapeIn<Args extends ShapeArgs>(args: Args, cont: Container) {
    const shape = createShape(args);
    cont.insert(shape);

    return shape;
}

const cont = new Container();
const circle2 = createShapeIn({type: "circle", radius: 1}, cont); // still inferred as Circle
const square2 = createShapeIn({type: "square", length: 1}, cont); // still inferred as Square

然而...

如果我再添加一种类型，一切都会崩溃。

让我们添加另一种类型的形状，称为 Figure:

interface FigureArgs { type: "figure", amount: number };
class Figure { constructor(_: FigureArgs) {}}

那么我们的联合类型将是:

type ShapeArgs = CircleArgs | SquareArgs | FigureArgs;
type Shape = Circle | Square | Figure;

最后，条件类型改为:

type ShapeOfArgs<Args extends ShapeArgs> = 
    Args extends CircleArgs? Circle :
    Args extends SquareArgs? Square :
    Figure;

现在这不再转译:

function createShape<Args extends ShapeArgs>(args: Args): ShapeOfArgs<Args> {
    switch (args.type) {
        case "circle": return new Circle(args); // error: Circle not assignable ShapeOfArgs<Args>
        case "square": return new Square(args); // error: Square not assignable ShapeOfArgs<Args>
        case "figure": return new Figure(args); // error: Figure not assignable ShapeOfArgs<Args>
    }
}

我做错了什么？

PS:我知道我可以使用 as 运算符强制它，但这违背了类型检查的目的。

TS Playground 链接:

• Good
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Answer 1

一个非常简单的解决方案是制作一个“类型字典”，而不是联合和条件类型:

// Dictionary of types
type ShapeByType = {
    circle: Circle,
    square: Square,
    figure: Figure
}

function createShape<Args extends ShapeArgs>(args: Args): ShapeByType[Args["type"]] {
    switch (args.type) {
        case "circle": return new Circle(args);
        case "square": return new Square(args);
        case "figure": return new Figure(args);
    }
}

const circle1 = createShape({type: "circle", radius: 1});
//    ^? Circle

// ...

const circle2 = createShapeIn({type: "circle", radius: 1}, container);
//    ^? Circle

有了这个，可以更容易地缩放到任意数量的形状。

Playground Link