Question

I have a makefile that can be reduced to this:

OBJS = obj1.o obj2.o

FLAGS = -Wall -Wextra -Werror -pedantic-errors -fno-rtti -std=c++0x
DEBUG_FLAGS = -ggdb -O0 -fstack-protector-all -D_GLIBCXX_DEBUG
RELEASE_FLAGS = -O3

release: $(OBJS)
    g++ $(FLAGS) $(RELEASE_FLAGS) $(OBJS)

debug: $(OBJS)
    g++ $(FLAGS) $(DEBUG_FLAGS) $(OBJS)

obj1.o: obj1.cpp
    g++ -c $(FLAGS) obj1.cpp

obj2.o: obj2.cpp
    g++ -c $(FLAGS) obj2.cpp

The problem is that all or none of the files must be built with the -D_GLIBCXX_DEBUG flag. I don’t know how to do this without writing two entries for every compilation unit, like

obj1_release.o: obj1.cpp
    g++ -c $(FLAGS) $(RELEASE_FLAGS) obj1.cpp

obj1_debug.o: obj1.cpp
    g++ -c $(FLAGS) $(DEBUG_FLAGS) obj1.cpp

How can I make the -D_GLIBCXX_DEBUG flag (and the other debug flags) take effect for all compilation units only when the user types make debug without writing two entries for every CU? (And vice versa; the release flags need to take effect on all CUs when the user types make release.)

I apologise if this is the basics of writing Makefiles, I don’t know much about them.

Answer 1

You are looking for pattern rules: something like this should do what you want. Note that this cannot be made to work correctly unless the debug and release versions of the program are given different names.

OBJS   := obj1 obj2 obj3
R_OBJS := $(OBJS:=_r.o)
D_OBJS := $(OBJS:=_d.o)

all: prog_r prog_d
release: prog_r
debug: prog_d

prog_r: $(R_OBJS)
    $(CXX) $(CXXFLAGS) $(RELEASE_FLAGS) $(LDFLAGS) $^ $(LIBS) -o $@

prog_d: $(D_OBJS)
    $(CXX) $(CXXFLAGS) $(DEBUG_FLAGS) $(LDFLAGS) $^ $(LIBS) -o $@

%_r.o: %.cc
    $(CXX) $(CXXFLAGS) $(RELEASE_FLAGS) -c $< -o $@

%_d.o: %.cc
    $(CXX) $(CXXFLAGS) $(DEBUG_FLAGS) -c $< -o $@

# header files
obj1_d.o obj1_r.o: foo.h bar.h
obj2_d.o obj2_r.o: quux.h
# ... etc ...