It is a long time ago, that I worked with GNU make, since then I forgot a lot. :-(
I have a project, containing some shell scripts and an extra directory with LaTeX documentation.
The diretory structure could look like this
.../topdir/
|
|-Makefile
|
|--data/
| |-filea
| |-fileb
|
|--latex/
| |-example.tex
| |-awesome-style.sty
| |-imagea.png
| |-Makefile
In that latex-directory exists a Makefile which is working alright, when I cd into the directory and call make clean or make example ...
Now, I want the Makefile in the topdir to be the master makefile. When I am in that directory and say e.g. make example the recipe should be, to change into the latex directory and run make from there, hence using latex/Makefile.
EDIT Of course, when I say make clean in the topdir Makefile, I want the it to also run make clean in the latex dir.
What is the correct trick? What must I write in the master makefile?
Just:
example:
$(MAKE) -C latex
clean:
$(MAKE) -C latex clean
.PHONY: example clean
will what you asked for. The .PHONY is optional but a good idea.
Related
I have a directory with test inputs and outputs. I wanted make to automatically test my program against this directory after build, for convenience. Thus I needed to somehow force the test target of Makefile to depend on the entire testing directory (it's called good, because it contains valid inputs and outputs for the program)
I read this question and the accepted answer and the comments about deleted files under this answer: Makefile rule that depends on all files under a directory (including within subdirectories) And, incorporating advice from this answer & comments, I came out with this:
my#comp:~/wtfdir$ cat Makefile
test : test.sh $(shell find good)
./test.sh
my#comp:~/wtfdir$
For the sake of MCVE, test.sh is very rudimentary:
my#comp:~/wtfdir$ cat test.sh
echo "blah"
my#comp:~/wtfdir$
However, I noticed, this behaves in a rather unexpected way:
my#comp:~/wtfdir$ ls good
test1 test1.out
my#comp:~/wtfdir$ make
./test.sh
blah
my#comp:~/wtfdir$ touch good/test1
my#comp:~/wtfdir$ make
cp good/test1 good/test1.out
./test.sh
blah
my#comp:~/wtfdir$
Why (expletive redacted) does modifying test1 cause make to overwrite test1.out with test1??? I'm not a big fan of data losses, you know.
What's going on here?
Your Make appears to be GNU Make. Here's why this happens. Your recipe:
test : test.sh $(shell find good)
./test.sh
adds to the prerequisites of test every file and directory that is listed
by find good in the current directory, which happen to be:
good
good/test1
good/test1.out
So to make target test, Make begins by determining if any of the specified
or built-in recipes require it to rebuild any of the prerequsities:
test.sh good good/test1 good/test1.out
Among its built-in recipes it finds:
%.out: %
# recipe to execute (built-in):
#rm -f $#
cp $< $#
as you can verify by running:
$ make --print-data-base | grep -A4 '%.out'
The rule for this recipe is matched by:
good/test1.out: good/test1
and by doing:
$ touch good/test1
you have made good/test1.out out of date with respect to good/test1.
So make executes the recipe:
#rm -f good/test1.out
cp good/test1 good/test1.out
the visible output of which is what you observed:
cp good/test1 good/test1.out
Then it proceeds with the recipe for test:
./test.sh
blah
There is always a risk of such booby-traps if you write a makefile that blindly
generates at runtime some set of preqrequisites or targets you don't know beforehand.
You could avoid this one in particular by explicitly deleting the offending
implicit pattern rule in your makefile by writing:
%.out: %
with no recipe. And you can avoid all possible booby-traps of this sort by disabling all
built-in recipes, with:
$ make --no-builtin-rules ...
but that will require you to write for yourself any builtin-recipes that your
makefile relies on.
The best solution for you is probably to amend your makefile as follows:
PREREQS := $(shell find good)
test : test.sh $(PREREQS)
./test.sh
$(PREREQS): ;
Then the last line explicitly specifies an empty recipe
for each of the $(PREREQS), and Make will not consult any pattern rules for targets
that have explicit recipes.
You should additionally make test a phony target:
.PHONY: test
for the avoidance of the booby-trap where something creates a file called test in the build directory.
I have my project binary located at my repository's root, along with a Makefile used to build it.
This binary uses many of my self-made libraries, located in my lib/ folder
For the purpose of building (and cleaning) my repository's binary, I want to implement the following execution :
Instead of hardcoding the following lines,
clean_binaries:
make -C clean lib/folder1 -s
make -C clean lib/folder2 -s
make -C clean lib/another_folder -s
I created the BIN_PATH variable, containing the previous paths.
BIN_PATHS = lib/folder1 \
lib/folder2 \
lib/another_folder
And made a simple rule like this one :
clean_binaries: $(BIN_PATHS)
make -C clean $< -s
BUT it only executes the line for the first field of the variable (lib/folder1), which is not what I want to do.
I thought about using implicit rules(?), just like I compile the .c files, but I couldn't get it right.
In the end, I simply wonder how to execute a rule for every field of a given variable, and this inside a Makefile, if there is any way to do so.
Thank you for your answers :]
The way you get GNU make to generate a sequence of commands that vary by the
fields in a variable is to use the foreach function, e.g.
Makefile
BIN_PATHS := lib/folder1 lib/folder2 lib/another_folder
.PHONY: clean_binaries
clean_binaries:
$(foreach path,$(BIN_PATHS),make -C $(path) clean ;)
which runs like:
$ make
make -C lib/folder1 clean -s; make -C lib/folder2 clean -s; make -C lib/another_folder clean -s;
not requiring a shell-loop.
Note also that you need to correct:
make -C clean <directory>
to:
make -C <directory> clean
I want make to remove all files except the source files and the make rule file (i.e. the file named makefile), so I added a phony rule at the end of my makefile:
.PHONY:clean
clean:
$(shell ls | grep -v "[.][ch]" | grep -v makefile | xargs rm)
This does what I intend. But make always complains
make: Nothing to be done for 'clean'.
After I run make clean. Why does this message appear? And how can I make it disappear?
The use of $(shell ...) is unnecessary. It runs the command, then the output is used as if it was part of the Makefile. There is no output, so the resulting rule is:
clean:
i.e. the actual list of commands to update the clean target is empty.
I have this make rule:
dist/node/%.js: src/%.node.js yarn.lock .babelrc | $(NODE_DIST_DIRS)
$(BIN)/babel $< -o $#
It works fine when my source file ends with .node.js. e.g.,
$ make dist/node/fs.js
mkdir -p dist/node/
node_modules/.bin/babel src/fs.node.js -o dist/node/fs.js
However, I want to compile dist/node/fs.js from either src/fs.node.js or src/fs.js, whichever exists.
The reason for this is that I have some shared files with just the .js extension (work in both browser and node), and then more specific files with the .node.js extension. If there's a more specific version, I want to use that.
I don't know how to do conditional dependencies in combination with %. Is this possible? Can I give precedence to dependencies and take the best match?
There is no way to do that in a single rule. You'll have to write two rules with different prerequisites but otherwise the same:
dist/node/%.js: src/%.node.js yarn.lock .babelrc | $(NODE_DIST_DIRS)
$(BIN)/babel $< -o $#
dist/node/%.js: src/%.js yarn.lock .babelrc | $(NODE_DIST_DIRS)
$(BIN)/babel $< -o $#
I suppose you could try using a double-colon rule, but those cannot be pattern rules.
As an alternative, consider generating and including a makefile containing just those dependencies.
I'm trying to modify a makefile to cross-compile binaries. The command in question is below:
# GNU Make solution makefile autogenerated by Premake
# Type "make help" for usage help
ifndef config
config=debug
endif
export config
PROJECTS := json openjaus openjaus-core openjaus-environment openjaus-mobility openjaus-manipulator openjaus-ugv Base Managed PingTest LargeMessageTest PdDemo GposDemo GposClientDemo StillImageSensorDemo StillImageClientDemo
.PHONY: all clean help $(PROJECTS)
all: $(PROJECTS)
json:
#echo "==== Building json ($(config)) ===="
#${MAKE} --no-print-directory -C .build -f json.make
As can be seen the makefile has several targets. They all have the same structure as the 'json' target. The command in question in question is
#${MAKE} --no-print-directory -C .build -f json.make
The '${MAKE}' variable = make (I have verified this with echo)
What does the -C do?
What does the .build do?
I'm good with -f json.make
Also, when I run make the json.make file gets created compiles file and deletes it self, so I do not have access to that file.
The error I receive when I modify the command in question is
==== Building json (debug) ====
make[1]: Nothing to be done for `/home/botbear/openwrt/trunk/staging_dir/toolchain- arm_v6k_gcc-linaro_uClibc-0.9.32_eabi/bin/arm-openwrt-linux-c++'.
The command after modifications looks like:
#${MAKE} /home/botbear/openwrt/trunk/staging_dir/toolchain-arm_v6k_gcc-linaro_uClibc-0.9.32_eabi/bin/arm-openwrt-linux-c++ --no-print-directory -C .build -f json.make
Any help is appreciated!
you can use man make to understand the parameters for make:
-C dir, --directory=dir
Change to directory dir before reading the makefiles or doing any-
thing else. If multiple -C options are specified, each is inter-
preted relative to the previous one: -C / -C etc is equivalent to
-C /etc. This is typically used with recursive invocations of
make.
-f file, --file=file, --makefile=FILE
Use file as a makefile.
so -C .build changes to the directory .build.
and i don't understand the part of your question about modifying the command.
Try to find where json.make lives. It seems that it's that makefile which creates & deletes the directory you were talking about.
From the command line it seems that make changes directory to .build and executes the json.make. Not sure how json.make ends up there. Is .build the directory which is created and then deleted?