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++
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