You need to use hasProperty here

done

what I meant is it that with this code

enum A { X }
enum A { Y }
let a = A.X | A.Y // A has both X and Y

you will see only X member, but you won't see Y since you are using one declaration to obtain property names.

Is there already a right way to get all members of enums? Otherwise I'll need to write getDeclarationsOfKind and then look at all members from those symbols.

well, you already have types on hand, so you can get its members that has flag SymbolFlag.EnumMember. Open question is if you need to pay attention to the order of member since it can directly affect values of enum members

I believe in the discussion around #1748 last week, we decided to ignore values of members. So as long as the enums are not const, we will just check for existence of matching property names.

The type object for enums don't actually have members defined, so there are no members to check for EnumMember.

symbol for a enum type should have exports that contain all enum members

resolveStructuredTypeMembers(enumType).members should do the trick

Sorry, given what Wesley said, I think it was correct to do what you were doing before when creating the map. If member.name isn't astring, how is this working? Don't you need member.name.text?

member.name is a string:

    function Symbol(flags: SymbolFlags, name: string) {
        this.flags = flags;
        this.name = name;
        this.declarations = undefined;
    }

Ah, it's the symbol, not the member itself. This is fine then.

lift it to a variable so it is not refetched on every iteration (cost of it is probably low but still)