Change definition of MemberExpr and MetaProp

[Austaras]

Well yes it's me again.

So once again spec allows a better definition fo AST leveraging rust enum, here is my proposed API

struct MemberExpr {
    pub span: Span,
    pub obj: Box<Expr>,
    pub prop: MemberProp,
}

enum MemberProp {
    Ident(Ident),
    PrivateName(PrivateName),
    Computed(ComputedPropName),
}

// add
struct SuperPropExpr {
    span: Span,
    prop: SuperProp
}

pub enum SuperProp {
    Ident(Ident),
    Computed(ComputedPropName),
}

struct MetaProp {
    span: Span,
    kind: MetaPropKind
}

enum MetaPropKind {
    NewTarget,
    ImportMeta
    // maybe add a no_exhaustive mark
}

However, this will likely be a great lot of works so I'm not sure if it worths it

---

New added: since we get rid of ExprOrSuper with MemberExpr, we could get better defintion for callee:

enum Call {
  Super, Import,
  Expr(Box<Expr>)
}
``

[kdy1]

I think it's a quite good change. MemberProp.obj being Expr was always a headache because of Expr::ident.

[Austaras]

Is it right that parse result of abc.foo = 123 is PatOrExpr::Pat(Box<Pat::Expr(Expr)>)?

[kdy1]

Honestly I'm not sure about PatOrExpr. I think Pat itself is enough, but it's lots of work.

[Austaras]

Seems to be caused by this line, what the purpose of it?

[kdy1]

Not all expressions are pattern, and some syntax is valid only for expressions.

e.g. foo() should not be Pat::Expr(). (That's why PatOrExpr exists)
At the same time, { a = b} is valid only for pattern, but we need AssignProp to be a variant of Prop because at the time of parsing we can't know if it's a pattern or an expression.