Allow inferring return type within a type guard in conditional types

Bug Report

🔎 Search Terms

typeguard conditional infer

🕗 Version & Regression Information

• This is the behavior in every version I tried, and I reviewed the FAQ for entries about (bug reporting)

⏯ Playground Link

Playground link with relevant code

💻 Code

This is simplified pretty heavily:

type InferredTypeGuard<TypeGuard> = 
    TypeGuard extends ((node: infer Input) => node is infer Output)
        ? (node: Input) => node is Output
        : never;

🙁 Actual behavior

Error on infer Output:

A type predicate's type must be assignable to its parameter's type.
  Type 'Output' is not assignable to type 'Input'.
    'Input' could be instantiated with an arbitrary type which could be unrelated to 'Output'.(2677)

🙂 Expected behavior

TypeScript should allow a conditional type to infer the Output type within the type guard function's return type.

There's no other syntax I can find to do so, and I would have thought this would be the cleanest syntax. Not sure if bug report or feature request...

从链接的 TypeScript Playground 示例和使用 infer Input 与 infer Output 的简化条件类型开始。跟踪报告错误 2677 的类型谓词可赋值性检查，并确定条件类型推断应如何与其交互。当示例能够使用预期推断出的输入类型和输出类型通过类型检查，并且为该行为提供回归覆盖时，即表示完成。