CFLAGS vs CPPFLAGS

Question

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.

You are after implicit make rules.