Question

对于我的应用程序，我使用了TypeScript manual中所述的带有完全检查性的“ Discriminated Union”模式。时间流逝，最终我的交换机最终包含了50多个案例。

所以我的问题是：有什么好的解决方案可以分解此开关，而又不会使其穷举？

换句话说，如何将其拆分，如果可以帮助我将这些并集逻辑上划分为子类型（例如，下面的形状可以划分为等边形和其他形状）：

interface Square {
    kind: "square";
    size: number;
}
interface Rectangle {
    kind: "rectangle";
    width: number;
    height: number;
}
interface Circle {
    kind: "circle";
    radius: number;
}

//... 50 more shape kinds

type Equilateral = Square | Circle /*| 25 more...*/;
type Other = Rectangle /*| 25 more...*/;

type Shape = Equilateral |  Other;

function assertNever(x: never): never {
    throw new Error("Unexpected object: " + x);
}
function area(s: Shape) {
    switch (s.kind) {
        case "square": return s.size * s.size;
        case "rectangle": return s.height * s.width;
        case "circle": return Math.PI * s.radius ** 2;
        /*
        ...
        ... a lot of code lines
        ...
        */
        default: return assertNever(s); 
    }
}

Answer 1

我刚刚发现（通过实验，不是因为在任何地方的文档中都提到了它），您确实可以使用多个判别式建立受歧视联合的类型层次结构：

interface Square {
    shape_kind: "equilateral";
    kind: "square";
    size: number;
}
interface Circle {
    shape_kind: "equilateral";
    kind: "circle";
    radius: number;
}
interface Rectangle {
    shape_kind: "rectangle";
    width: number;
    height: number;
}

type Equilateral = Square | Circle

type Shape = Equilateral | Rectangle;

function area(s: Shape) {
    switch (s.shape_kind) { // branch on "outer" discriminant
        case "equilateral":
            // s: Equilateral in here!
            return area_root(s) ** 2;
        case "rectangle":
            return s.height * s.width;
    }
}
function area_root(e: Equiliteral) {
    switch (s.kind) { // branch on "inner" discriminant
        case "square": return s.size;
        case "circle": return Math.sqrt(Math.PI) * s.radius;
    }
}

如何分解TypeScript“ Discriminated Union”开关块并同时使其详尽无遗

1 个答案: