How to get only file name in preprocessor?

Question

I am (was) using the __FILE__ and __LINE__ macros for printing diagnostic messages out of my code. This works quite well when you use GCC with make, the file is as short as you specified it on the command line. I recently switched to using CodeLite which uses fully qualified file names (at least under windows) when building. Suddenly my diagnostic output is almost not readable.

It there a way to get only the file component of the filename in the preprocessor? I can live with a non portable GCC specific solution. (I will fallback to plain __FILE__ other cases.)

Sure I can pass the contents of __FILE__ through a function and extract only the file component, but string operations was not what I had in mind for diagnostic messages that should not change runtime behavior...

NOTE: I use the filename the way GNU uses it. A Path is collection of filenames and a filename is either a relative or absolute identifier of a file. A filename can be made up of a directory component and file component.

Answer 1

If you are using GNU Make then you can simply pass -D BASE_FILE_NAME=\"$*.c\" in on the preprocessing stage of compilation (if you're doing them separately, or at compilation if in a single stage, which is the norm).

This depends upon the way you have your file names determined. Mine come from a list of plain file names and are prefixed with directories using functions in the makefile at a later stage.

IE, this works well for me, but your mileage may vary! :-)

A simplified version of my make "code" :

CLASSES = main.c init.c

PREPROCESSED = $(patsubst %.c,$(PPCDIR)/%.pp.c,$(CLASSES))

$(PREPROCESSED): $(PPCDIR)/%.pp.c: %.c $(ALLH) 
    $(GCC) $(GCCOPTS) -D BASE_FILE_NAME=\"$*\" -E $< > $@

The simply use BASE_FILE_NAME in your code as you like :-)

Answer 2

There is no known preprocessor macro that provides the functionality. Passing __FILE__ through a function seams like the only sensible option.