Why is the return type for the always true conditional not passing the type check in this scenario?

Upon examination,

type B = { 
  foo: string;
  bar: number;
};

function get<F extends B, K extends keyof B>(f: F, k: K): F[K] {
  return f[k];
}

It seems like a similar concept is expressed in a different way in the following code snippet:

function get<F extends B, K extends keyof B>(f: F, k: K): F extends B ? F[K] : undefined {
  return f[k];
}

However, an error occurs during type checking:

Type 'F[K]' is not assignable to type 'F extends B ? F[K] : undefined'.
  Type 'F["foo"] | F["bar"]' is not assignable to type 'F extends B ? F[K] : undefined'.
    Type 'F["foo"]' is not assignable to type 'F extends B ? F[K] : undefined'.
      Type 'string' is not assignable to type 'F extends B ? F[K] : undefined'.

This raises a question about what additional information the TS compiler has about types that may not be immediately obvious.

Playground: http://www.typescriptlang.org/play/#code/C4TwDgpgBAQlC8UDeUBQUoDMD22BcUAzsAE4CWAdgOYDc6UARgIYkEUCuAtgxCXQL51M7CgGNgZbBShUIwADwAxKBAAewCBQAmhWABooAaRXrNOqAGsIIbJlgA+ABSYCigxYKGAlK5MbtunAA-FCKANqGALpQBCJaEJiUEFrI9CRy7CTSmGEWkQKoQA

Answer №1

It's as if you possess a greater understanding than the compiler itself... The compiler does not delve deeply into analyzing conditional types that rely on unresolved generic type parameters. This means that the conditional type F extends B ? F[K] : undefined is not actually assessed within the function implementation, especially when F has not been assigned a specific type yet. Even though it may seem obvious to you that F extends B ? F[K] : undefined equals F[K], due to the equivalence of the condition F extends B and the generic constraint F extends B, the compiler refrains from executing this process (most likely for performance optimization purposes).

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