$ 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
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