My Makefile looks like this:
%.foo: %.bar cp $< $@ test: *.foo echo *.foo I have 2 files in the directory: a.bar and b.bar. When I run make test it outputs:
cp b.bar *.foo echo *.foo *.foo It also creates a file *.foo in the current directory.
I am actually expecting to see this:
cp a.bar a.foo cp b.bar b.foo echo *.foo a.foo b.foo And also creating a.foo and b.foo. How to achieve that?
In this case you need to handle wildcards explicitly, with the wildcard function (at least in GNU Make):
%.foo: %.bar cp $< $@ foos = $(patsubst %.bar,%.foo,$(wildcard *.bar)) test: $(foos) echo $(foos) $(wildcard *.bar) expands to all the files ending in .bar, the patsubst call replaces .bar with .foo, and all the targets are then processed as you’d expect.
There is no *.foo file to begin with. So what make does is look for how to make *.foo literaly and the first rule does this. Make expands $< to the first pre-requisite (*.bar, which happens to be b.bar in this case). Make then runs the shell command cp b.bar *.foo. Since there is no *.foo, shell expands it to cp b.bar *.foo literally. That's how you get a *.foo file.
You can verify this by running make -d test.
You can get the effect you want by generating the list of targets based on list of prerequisites.
TARGETS = $(patsubst %.bar,%.foo,$(wildcard *.bar)) %.foo: %.bar @cp $< $@ test: $(TARGETS) @echo $(TARGETS) echo *.foo 