[makefile] 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++