CFLAGS vs CPPFLAGS
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++
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.