能否使用mappend/条件类型扩展并转换判别式联合类型？

Absolutely! You can pull off this transformation using TypeScript conditional types—this is essentially the type-level equivalent of a "mappend" operation for your discriminated union. Let's build the DUTransformer type step by step to get exactly the result you want.

Step 1: Restate your original discriminated union #

First, let's lay out your base type for clarity:

type MyDU = | { kind: 'foo' } | { kind: 'bar' };

Step 2: Define the DUTransformer type #

We'll leverage distributed conditional types (a TypeScript feature where conditional types automatically apply to each member of a union) to extend each union member with the matching boolean property:

type DUTransformer<T extends { kind: string }> = T extends infer U
  ? U & Record<U['kind'], boolean>
  : never;

Step 3: Check the transformed result #

When we apply this transformer to MyDU, we get exactly the union you're targeting:

type Transformed = DUTransformer<MyDU>;

// This resolves to:
// type Transformed = 
//   | { kind: 'foo'; foo: boolean } 
//   | { kind: 'bar'; bar: boolean }

How this works under the hood #

Let's break down the logic for each union member:

• The conditional T extends infer U triggers TypeScript's distributed behavior, splitting MyDU into its individual parts ({ kind: 'foo' } and { kind: 'bar' }).
• For each member U, U['kind'] extracts the literal string value from the kind property (either 'foo' or 'bar').
• Record<U['kind'], boolean> creates a new type with a single property matching that literal string, typed as boolean (e.g., { foo: boolean }).
• The & operator merges the original union member with this new property type, creating the extended object type.
• Finally, TypeScript automatically recombines these extended types back into a single discriminated union.

If you want to make the transformer more flexible (e.g., support kind values that are numbers or symbols), you can adjust the constraint to T extends { kind: PropertyKey } instead of string.

内容的提问来源于stack exchange，提问作者robkuz