Re: Dropping "call" operator?

On Tue, 2024-03-26 at 10:15 -0700, Kaz Kylheku wrote:
> However, I'm noticing that expansion preserves the semantic
> boundaries, which is important. If we have $(x) that expands
> to a,b, then we /can/ invoke $(call $(x),c,d). The delimiting
> of argument material must be happening prior to expansions
> (at least in the abstract).

Yes, this is true for most parts of the parser and is a common and
long-standing way to avoid issues with special characters and parsing.
There's no chance of this ever changing.

Make (almost) always does word separation BEFORE it does expansion.  I
actually can't think of a place where this is not true but it's
possible I'm forgetting something.