Re: ssh, $(MAKE) and -n weirdness

[Martin d'Anjou]

> > $ cat makefile
> > remote=localhost
> > remote1:
> >        ssh $(remote) "cd $(PWD) && $(MAKE) -f makefile this"
> > remote2:
> >        ssh $(remote) "cd $(PWD) && make -f makefile this"
> > this:
> >        @echo This is done!
> >
> > $ make remote1 -n
> > ssh localhost "cd /home/me && make -f makefile this"
> > This is done!
> > $ make remote2 -n
> > ssh localhost "cd /home/me && make -f makefile this"
> > $
> >
> > What I do not understand is why "remote1" is built even though I pass -n?
>
> This is documented. See GNU Make Manual section 5.7.1: "Command lines
> containing MAKE are executed normally despite the presence of a flag that
> causes most commands not to be run".

True. I wanted to point out that in building remote1, the -n flag is not 
passed to the submake, but now I understand that it is due to the fact it 
is in the ssh subshell:

  $ make remote1 -n
  ssh localhost "cd /home/me && make this"
  This is done!

Instead of this:

  $ make remote1 -n
  ssh localhost "cd /home/me && make this"
  echo This is done!

However, if I have this makefile instead:

  $ cat makefile
  remote=localhost
  remote1:
         ssh $(remote) "cd $(PWD) && $(MAKE) -f makefile this MAKEFLAGS=$(MAKEFLAGS) 
MAKEOVERRIDES=$(MAKEOVERRIDES)"
  this:
         @echo This is done!

Then I get the desired effect of propagating the '-n' and command line
variables through the subshell (ssh in this case). I am sure this will 
come back to bite me, but I don't see how yet.

Martin