Cannot infer `thisArg` argument type of `Function.bind` for functions with a `this` type consisting of generic functions

Bug Report

Consider a function like-

declare function callback<T>(this: CallbackCtx<T>): void;

Where CallbackCtx is an object containing generic function(s)-

interface CallbackCtx<T> {
    onsuccess: (x: T) => number;
    onerror: (err: any) => void;
}

For such a function, that has its this type set to an object containing generic function(s), Function.bind cannot infer the type of thisArg (from the first overload) correctly.

If the type parameter to .bind is explicitly specified (by substituting the generic function type to be a concrete type) - it works fine.

The type of thisArg inferred by typescript automatically, is CallbackCtx<unknown>.

It should be noted that if CallbackCtx contains only non function-like property(s) - the types are inferred correctly. As seen in this playground

🔎 Search Terms

bind, generic, function

🕗 Version & Regression Information

4.2.3

⏯ Playground Link

Playground link with relevant code

💻 Code

interface CallbackCtx<T> {
    onsuccess: (x: T) => number;
    onerror: (err: any) => void;
}

declare function callback<T>(this: CallbackCtx<T>): void;

declare const ctx: CallbackCtx<string>;

callback.bind(ctx);
// ^ CallbackCtx<string> is not assignable to CallbackCtx<unknown>

callback.bind<(this: CallbackCtx<string>) => void>(ctx);
// ^ Works fine when manually substituting the generic type of the function

🙁 Actual behavior

The line callback.bind(ctx); produces errors

🙂 Expected behavior

The line callback.bind(ctx); should infer the type of callback correctly by substituting the generic type with the help of the type of ctx and should not produce errors

Start with the linked TypeScript Playground and the callback.bind(ctx) example, comparing it with the explicitly typed .bind call. Trace how the this type containing generic function properties is inferred for Function.bind. Done means the inferred context uses CallbackCtx<string> from ctx and the unannotated call produces no error.