filter-out on shell output - makefile

Question

There are four folders in my src directory: dir1, dir2, dir3, dir4.
PARSE_MODULES = $$(MODULES=$$(ls -l $1 | grep '^d' | awk '{print $$9}'); echo $$MODULES);
MODULES:=$(call PARSE_MODULES,src)
all:
#echo $(MODULES)
Hence, this little piece of code outputs
dir1 dir2 dir3 dir4
Now, I add the following two lines before the all rule:
X:=dir1 dir3
MODULES:=$(filter-out $(X),$(MODULES))
The output remains unchanged, i.e., filter-out is not working as expected. However, if I manually define MODULES, I get the expected output:
MODULES:=dir1 dir2 dir3 dir4
X:=dir1 dir3
MODULES:=$(filter-out $(X),$(MODULES))
Output: dir2 dir4
Why is it impossible to process the output of awk in filter-out? Is there any way to do so?

Answer 1

Your problem is that you have deferred expansion of the contents of PARSE_MODULES by using $$, so that it's being run by the shell in the recipe, not by make.
This:
PARSE_MODULES = $$(MODULES=$$(ls -l $1 | grep '^d' | awk '{print $$9}'); echo $$MODULES);
MODULES:=$(call PARSE_MODULES,src)
leaves the MODULES variable containing this text:
$(MODULES=$(ls -l src | grep '^d' | awk '{print $9}'); echo $MODULES);
Then when you use $(MODULES) in the recipe, it expands like this:
all:
#echo $(MODULES=$(ls -l src | grep '^d' | awk '{print $9}'); echo $MODULES);
then the shell will run the command inside $(...) and echo the results.
When you try to filter out names like dir1 etc. from the value of the make variable MODULES, nothing happens because the make variable MODULES contains the shell script itself, as text, it doesn't contain the results of executing the shell script.
I'm not sure why you are trying to do things in such an incredibly convoluted way, with so much escaping and using difficult to understand shell scripting. Maybe there's a reason that your full environment needs something so complicated but you can FAR more easily implement the example above just using some basic make functions:
PARSE_MODULES = $(sort $(notdir $(patsubst %/.,%,$(wildcard $1/*/.))))
MODULES := $(call PARSE_MODULES,src)
Now the PARSE_MODULES is expanded by make, not by the recipe, and filter-out etc. will work fine on it. Plus it's a lot simpler then doing a shell with grep and awk.

Answer 2

As far as I understand the shell list $(MODULES=$(ls -l src | grep '^d' | awk '{print $9}'); echo $MODULES) is evaluated by the shell when the recipe of all is executed. This is too late. Try this, maybe:
PARSE_MODULES = $(shell ls -l $1 | grep '^d' | awk '{print $$9}')
MODULES := $(call PARSE_MODULES,src)
X := dir1 dir3
MODULES := $(filter-out $(X),$(MODULES))
all:
#echo $(MODULES)

Answer 3

Minor simplification from OP version:
# possibly: find -maxdepth 1
PARSE_MODULES = $(shell find $1 -type d -printf "%P\n")
SKIP := dir1 dir3
MODS := $(call PARSE_MODULES,src)
FLTR := $(filter-out $(SKIP), $(MODS))
all:
#echo MODS: $(MODS)
#echo SKIP: $(SKIP)
#echo FLTR: $(FLTR)
Output:
MODS: dir1 dir2 dir3 dir4
SKIP: dir1 dir3
FLTR: dir2 dir4