Makefile patsubst using value of shell variable - linux

Question

As part of a make install rule for a testing suite, I'd like to move all binary executables in one directory (a src directory) to a bin directory. I thought an easy way to do this would be to simply loop over each file in the src directory and then use patsubst to replace src with bin in each path. Unfortunately, I can't get it to work because I can't get make to evaluate the name of the current FILE in each loop iteration. All I have access to is the bash shell variable $$FILE, but when I use this with the make patsubst function, it doesn't actually evaluate the shell variable $$FILE... rather, the patsubst function seems to just see the string "$FILE".
So, here is what I'm trying:
install :
-- irrelevant stuff snipped --
for FILE in $(BINARY_TARGETS); do \
if [ -f $$FILE ]; then mv -f $$FILE $(patsubst %/src/,%/bin/,$$FILE); fi \
done
This results in an error for each file:
mv: ‘./src/foo/bar’ and ‘./src/foo/bar’ are the same file
This error leads me to understand that the patsubst function in make is not actually evaluating shell variables, but just sees $FILE, and so the result is that it doesn't find the substitution pattern, and the final command passed to mv has the source and destination path as the same string.
So, is there a way to get patsubst to evaluate the value of a shell variable? Or is there a better way in general to accomplish what I'm trying to achieve here?

Answer 1

make processing has a precedence over passing commands to shell. And, once passed, they are executed by shell. So, at first make, processes the command and in:
$(patsubst %/src/,%/bin/,$$FILE)
$$FILE is substituted by $FILE and then treated literally. So, no pattern is matched and in effect patsubst returns $FILE. Please see following example:
bar:
echo $(patsubst %/src/,%/bin/,$$whatever)
It gives:
arturcz#szczaw:/tmp/m$ make bar
echo $whatever
arturcz#szczaw:/tmp/m$
As a result of your makefile rule bash is given following command to execute:
for FILE in src/a src/b src/c; do \
if [ -f $FILE ]; then mv -f $FILE $FILE; fi \
done
and that's why you got your result.
Solution
You can rely on bash to do a proper substitution, but you need to enforce it as a shell (by default it is sh, which lacks some required features):
SHELL=bash
install:
for FILE in $(BINARY_TARGETS); do \
if [ -f $$FILE ]; then echo $$FILE $${FILE/\/src\//\/bin\/}; fi \
done
You can also ask make to do a loop and substitution. There are few ways you can achieve that. This one is doing all the replacement and prepares command on the fly.
install:
$(foreach d,$(BINARY_TARGETS),if [ -f $(d) ]; then mv -f $(d) $(d:./src/%=./bin/%);fi;)
You can cease checking existence of files to make too by using `$(wildcard) function:
install:
$(foreach d,$(wildcard $(BINARY_TARGETS)),mv -f $(d) $(d:./src/%=./bin/%);)
And, finally, solution which I personally prefer - do it in a make way using a proper dependencies and rules.
install: $(BINARY_TARGETS:./src/%=./bin/%)
bin/%: src/%
mv -f $< $#
If existence any of files in BINARY_TARGET is optional, you may want to use the $(wildcard) trick again:
install: $(patsubst ./src/%,./bin/%,$(wildcard $(BINARY_TARGETS)))
bin/%: src/%
mv -f $< $#