I have an external program creating the a bunch of files. These files are prerequisites to my targets. At the moment I call it as follows:
# build the external resouce and create a the file filelist.txt
VHDL_SOURCES += $(shell make -C $(SCRIPT_PATH) > $(SCRIPT_PATH)/log; cat $(BUILD_PATH)/filelist.txt)
This works fine. My variable VHDL_SOURCES contains all source files mentioned in filelist.txt. The drawback of this approach is that I can not see the output of the called script. As it takes a long time to run, it would be great to see on the stdout whats going on.
Is there a way to show what is moved (and therefore hidden) to $(SCRIPT_PATH)/log at the moment?
question from:https://stackoverflow.com/questions/65881794/show-the-output-of-a-shell-xx-command-also-on-stdout