On Sun, 2024-01-28 at 18:06 -0500, rsbec...@nexbridge.com wrote:
> > FOO +:= bar
> >
> > can be interpreted as working like this:
> >
> > FOO := $(FOO) bar
> >
> > which is what you and others are arguing for. Or it can be
> > interpreted as working
> > like this:
> >
> > __FOO :=
On Sunday, January 28, 2024 5:36 PM, Paul Smith wrote:
>On Sat, 2024-01-27 at 17:45 -0500, rsbec...@nexbridge.com wrote:
>> My take on it is that +:= (because of the : ) means that you have to
>> resolve everything at that point.
>
>Yes, I understand what you are saying. The question is, is that t
