utility-types | utility types , complementing TypeScript built | Functional Programming library

You can use interfaces instead:

So, the answer is that T is not equal with A. It means that T can be another type, that can do something and implements keys from A if you want them to be equal you need to do this.

By merging the [cmp] method into the global Function interface:

You probably want a type intersection:

You can flatten a union using StrictUnion as defined here This type will basically add the missing members to all union constituents with the type undefined. This will allow de-structuring to suggest all members from any constituent, but each member that is not present in all union constituents will also contain undefined (which is probably for the best from a type-safety perspective)

The most important piece you need is to identify the optional properties of an object type programmatically. Here's one way to do it (as long as your object type doesn't have index signatures):

TL;DR: It’s impossible to convert a tuple type into an object, since information about the key is missing from the tuple.

When you say you have a tuple type like [session: SessionAgent, streamID: string], I guess you really mean [SessionAgent, string].

You don’t get to keep the variable names along side the tuple, they’re discarded, and there’s no way to restore lost information.

A workaround, if it suits you, would be converting MyClass constructor signature from positional params to named params.

I've often been bothered by that reuse of extends too, but it's true in a way. The type is "a" | "b" | "c", but you could use something else assignable to that, such as "a" | "b":

It would be really nice if Typescript could infer this automatically, but it's not so automatic. It does give you a way to specify it explicitly using overloads though.

Overloads are a way to define the return type for different variations of a function signature. For a generic async lambda like in your example, this looks like: