pipe with no arguments should return an identity function type?

[lffg]

Type inference "bug" report

I'm new to this library, and I don't actually know if this is a by-design decision.

Current Behavior

When we use pipe with no arguments, an identity function is returned, as we can see here:

rxjs/src/internal/util/pipe.ts

Line 25 in fd66c29

return identity as UnaryFunction<any, any>;

The problem is that the current type does not reflect this behavior, see (in the 5th line of that same file):

rxjs/src/internal/util/pipe.ts

Line 5 in fd66c29

export function pipe<T>(): UnaryFunction<T, T>;

I think that, if we change that first overload to:

export function pipe(): typeof identity;

// Or:
export function pipe(): <T>(x: T) => T;

Diff

- export function pipe<T>(): UnaryFunction<T, T>; 
+ export function pipe(): typeof identity;

Or:

- export function pipe<T>(): UnaryFunction<T, T>; 
+ export function pipe(): <T>(x: T) => T;

The type inference would be more accurate, see:

Current type inference

// %inferred-type: UnaryFunction<unknown, unknown>
const fn = pipe();

// %inferred-type: unknown
const returnValue = fn('Foo');

Expectedtype inference

// %inferred-type: <T>(x: T) => T        /* (identity function type) */
const fn = pipe();

// %inferred-type: "Foo"
const returnValue = fn('Foo');

Environment

• RxJS version: Latest (6.6.0 at the moment)

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Should I open a pull request to fix this? Or is this behavior intentional?