> it's not yet clear to the parser

Not at all. C/C++ grammar is explicitly contextual. `foo * bar` is a declaration if-and-only-if `foo` was previously declared as a type. Otherwise it is an expression. The subsequent tokens have zero bearing on this.

Nor is this a good argument for types-to-the-right. If C named pointer types like `⋆foo` instead of `foo⋆` [stars used to avoid formatting glitches], then putting the type to the right as `bar * foo` would be equally ambiguous as your example.