Re: Multiple rules for same target: what is the order of prerequisites in $^ and what does $< expand to?

I apologize for using the wrong automatic variable. I had the firm
assumption that we were talking about "$^", not "$<" as I would never use
the latter in a case where the genesis of its content is not explicit.

Am Mo., 2. Okt. 2023 um 15:08 Uhr schrieb Paul Smith <[email protected]>:

> On Mon, 2023-10-02 at 09:06 +0200, Mark Piffer wrote:
> > For the sake of reliability I wouldn't use pure `$<` anywhere.
>
> I'm not sure what you mean by "reliability" (I'm not aware of any
> change to the behavior of $< in the 30+ year history of GNU make) and I
> don't know what you mean by "pure $<"; what would one use instead, that
> is "non-pure"?
>
> > And if the answer to your question "Which one, listed in another
> > rule, becomes $<?" isn't "All collected prerequisites, with
> > duplicates purged"
>
> I don't understand what this means.  $< is always a single word.  It's
> never multiple words so it can never be "all collected prerequisites".
>
> Also make doesn't drop duplicates (this is necessary because sometimes,
> for example for linking, duplicates are important).  But that doesn't
> matter for $< which is always a single word, as I mentioned.
>
> The behavior is straightforward: when make parses a rule _without_ a
> recipe, it will append new prerequisites to the END of its targets'
> list of prerequisites.
>
> When make parses an explicit rule that _has_ a recipe, then those
> prerequisites are added to the BEGINNING of the target's list of
> prerequisites.
>
> For implicit rules, the prerequisites on the matching rule are added to
> the BEGINNING of the target's list of prerequisites when it matches.
>
> Then, when $< is expanded, it expands to the first prerequisite.
>