如何在 typescript 4.5.4 中定义方法返回类型的映射类型？

Question

In typescript 4.4.4 this code compiled fine:

/**
 * type to get only those properties that are functions
 */
type FunctionProperties<T> = {
  [P in keyof T]: T[P] extends (...args: any) => any ? P : never;
}[keyof T];

type ReturnTypeOfMethod<T, K extends FunctionProperties<T>> = ReturnType<T[K]>;

see Playground 4.4.4 example

But the same code fails to compile in typescript 4.5.4 - see Playground 4.5.4 example

Type 'T[K]' does not satisfy the constraint '(...args: any) => any'.
  Type 'T[FunctionProperties<T>]' is not assignable to type '(...args: any) => any'.
    Type 'T[T[keyof T] extends (...args: any) => any ? keyof T : never]' is not assignable to type '(...args: any) => any'.
      Type 'T[keyof T]' is not assignable to type '(...args: any) => any'.
        Type 'T[string] | T[number] | T[symbol]' is not assignable to type '(...args: any) => any'.
          Type 'T[string]' is not assignable to type '(...args: any) => any'.

The strange thing is that the ReturnTypeNumber type still shows the expected type ( number in this example).
Any idea why this does not work anymore or how to fix this (using @ts-ignore is not a fix)?

Answer 1

Turns out that this was a documented breaking change in version 4.5:

Restrictions on Assignability to Conditional Types TypeScript no longer allows types to be assignable to conditional types that use infer, or that are distributive. Doing so previously often ended up causing major performance issues. For more information, see the specific change on GitHub.

So we really need the more verbose version as mentioned in Stonehearts answer

Answer 2

Quite verbose since you have to repeat the condition, but it would suppress the error.

type ReturnTypeOfMethod<T, K extends FunctionProperties<T>> =
  T[K] extends (...args: any) => any ? ReturnType<T[K]> : never;