推断 typescript 通用包装器 function 返回类型属性值

Question

How to properly infer xxx type to data ?

Restrictions:

• wrapper argument should accept only Promise<{ xxx: Data; }> Promise<{ xxx: Data; }> - currently works.
• only wrapper can be changed

Typescript sandbox

function wrapper<
    Data,
    A extends Promise<{ xxx: Data; }>,
    >(a: A): Promise<{ data: Data }> {
    return 1 as any
}


async function a(): Promise<{ xxx: string }> {
    return { xxx: 'a' }
}


wrapper(a()).then(res => {
    const data: string = res.data // is unknown, how to make it infer string?
})

Answer 1

Presumably you expect the compiler to infer A as Promise<{ xxx: string }> , and from there to infer Data as string because A extends Promise<{xxx: Data}> . Unfortunately, in TypeScript, generic constraints like extends Promise<{xxx: Data}> don't serve as inference sites for type parameters. (Such behavior was suggested in microsoft/TypeScript#7234 but was never implemented.) So the compiler has no idea what to infer Data as, and thus it falls back to the unknown type . Oops.

In order to get inference for Data (which I will rename to D to be more in keeping with the usual naming conventions for type parameters) from a , you'll need to give a a type which is directly related to D . Here's the easiest way to do it:

function wrapper<D>(a: Promise<{xxx: D}>): Promise<{ data: D }> {
    return 1 as any
}

You weren't really using A in your code, so I left it out. Instead, we say that a is of type Promise<{xxx: D}> . The compiler can infer D from that, by matching Promise<{xxx: string}> to Promise<{xxx: D}> .

Let's see it in action:

wrapper(a()).then(res => {
    const data: string = res.data // okay
});

Looks good.

Playground link to code