Description

declare const f:
  <T extends F<T>>(t: T) => T

type F<T> =
  { a: unknown
  , b: (a: T extends { a: infer X } ? X : never) => unknown
  }

f({
  a: "hello",
  b: x => x.toUpperCase()
})
// doesn't compile because infers `x` as `unknown` instead of `string`

f({
  a: "hello",
  b: (x: string) => x.toUpperCase()
})
// compiles

Contributor guide

Tech stack

typescript

Domain

developer experience

Issue type

bug

DifficultyEstimated implementation difficulty for a new contributor, from 1 for very small changes to 5 for expert-level work.

5

Estimated timeA rough time range for an experienced contributor to investigate, implement, test, and prepare a pull request.

over 1 week

Activity statusHow available the issue appears right now: fresh, active, stale, blocked, or waiting on maintainer input.

stale

ClarityHow clearly the issue explains the expected change, acceptance criteria, and next step.

clear

Prerequisites

TypeScript genericsType inference internals

Newbie friendlinessA 1-100 score estimating how approachable this issue is for first-time contributors.

10

Research direction

The issue demonstrates a failure in type inference for a circular constraint T extends F<T>. To resolve, investigate the type inference algorithm in the TypeScript compiler, particularly how constraints and conditional types interact. Look at the provided playground link for reproducing the issue. Also review related issues #40439 and #30134 for context. The fix may involve modifying the constraint instantiation logic in the checker.