在 WSL 下的 Makefile 中运行带有标志的 echo

Question

I am trying to run a Makefile from under WSL that contains the following lines:

debug: create_soft_links
@mkdir -p Debug64
@echo -e 'all: bld'                                             > Debug64/Makefile
@echo                                                          >> Debug64/Makefile
@echo -e '%.o: ../../%.c'                                      >> Debug64/Makefile
@echo -e '\tgcc -g $$(CFLAGS) $$(INCLUDE) $$< -o $$@'          >> Debug64/Makefile

Problem is that the resulting Debug64/Makefile file looks like this:

-e all: bld

-e %.o: ../../%.c
-e      gcc -O3 $(CFLAGS) $(INCLUDE) $< -o $@

A colleague just showed me on an actual Linux machine that the make command works correctly there, and the preceding -e flag is not printed in the generated Debug64/Makefile . What am I doing wrong?

Answer 1

Use instead of echo the printf(1) command. So your last line would be

@printf "\tgcc -g %s %s $$< -o $$@\n" $$(CFLAGS) $$(INCLUDE)

BTW, if you generate your build automation script, consider switching to ninja . You might use Guile or Python or GNU awk as such a generator.