Re: The first word in $(MAKEFLAGS) is not a group of short options
Paul Smith <[email protected]>
| Newsgroups | gmane.comp.gnu.make.bugs |
|---|---|
| Organization | GNU's Not UNIX! |
| Message-ID | <[email protected]> |
On Mon, 2025-02-10 at 02:01 +0900, Masahiro Yamada wrote:
> The first word in $(MAKEFLAGS) is not a group of short options.
>
> $(MAKE) MAKEFLAGS=--no-print-directory SUBMAKE=1
If you override the value of MAKEFLAGS as a command-line variable
assignment, then the value you specified is the value you'll get.
Maybe. Most of the time.
You should never add MAKEFLAGS to the command line as a variable
override like this. You should use:
$(MAKE) --no-print-directory SUBMAKE=1
then make will correctly construct the MAKEFLAGS variable for you.