I understand that CFLAGS (or CXXFLAGS for C++) are for the compiler, whereas CPPFLAGS is used by the preprocessor.
But I still don't understand the difference.
I need to specify an include path for a header file that is included with #include -- because #include is a preprocessor directive, is the preprocessor (CPPFLAGS) the only thing I care about?
Under what circumstances do I need to give the compiler an extra include path?
In general, if the preprocessor finds and includes needed header files, why does it ever need to be told about extra include directories? What use is CFLAGS at all?
(In my case, I actually found that BOTH of these allow me to compile my program, which adds to the confusion... I can use CFLAGS OR CPPFLAGS to accomplish my goal (in autoconf context at least). What gives?)
The implicit make rule for compiling a C program is
%.o:%.c
$(CC) $(CPPFLAGS) $(CFLAGS) -c -o $@ $<
where the $() syntax expands the variables. As both CPPFLAGS and CFLAGS are used in the compiler call, which you use to define include paths is a matter of personal taste. For instance if foo.c is a file in the current directory
make foo.o CPPFLAGS="-I/usr/include"
make foo.o CFLAGS="-I/usr/include"
will both call your compiler in exactly the same way, namely
gcc -I/usr/include -c -o foo.o foo.c
The difference between the two comes into play when you have multiple languages which need the same include path, for instance if you have bar.cpp then try
make bar.o CPPFLAGS="-I/usr/include"
make bar.o CFLAGS="-I/usr/include"
then the compilations will be
g++ -I/usr/include -c -o bar.o bar.cpp
g++ -c -o bar.o bar.cpp
as the C++ implicit rule also uses the CPPFLAGS variable.
This difference gives you a good guide for which to use - if you want the flag to be used for all languages put it in CPPFLAGS, if it's for a specific language put it in CFLAGS, CXXFLAGS etc. Examples of the latter type include standard compliance or warning flags - you wouldn't want to pass -std=c99 to your C++ compiler!
You might then end up with something like this in your makefile
CPPFLAGS=-I/usr/include
CFLAGS=-std=c99
CXXFLAGS=-Weffc++
You are after implicit make rules.
The CPPFLAGS macro is the one to use to specify #include directories.
Both CPPFLAGS and CFLAGS work in your case because the make(1) rule combines both preprocessing and compiling in one command (so both macros are used in the command).
You don't need to specify . as an include-directory if you use the form #include "...". You also don't need to specify the standard compiler include directory. You do need to specify all other include-directories.
To add to those who have mentioned the implicit rules, it's best to see what make has defined implicitly and for your env using:
make -p
For instance:
%.o: %.c
$(COMPILE.c) $(OUTPUT_OPTION) $<
which expands
COMPILE.c = $(CXX) $(CXXFLAGS) $(CPPFLAGS) $(TARGET_ARCH) -c
This will also print # environment data. Here, you will find GCC's include path among other useful info.
C_INCLUDE_PATH=/usr/include
In make, when it comes to search, the paths are many, the light is one... or something to that effect.
C_INCLUDE_PATH is system-wide, set it in your shell's *.rc.$(CPPFLAGS) is for the preprocessor include path.VPATH = my_dir_to_search
... or even more specific
vpath %.c src
vpath %.h include
make uses VPATH as a general search path so use cautiously. If a file exists in more than one location listed in VPATH, make will take the first occurrence in the list.
Source: Stackoverflow.com