Way to omit undefined preprocessor branches by default with unifdef?

Question

I'm using a complicated C code that includes many, many compilation options. This makes the code very hard to read. I'd like to produce a copy of the code reflecting the way it's actually compiled. I've gotten pretty good results using the "unifdef" utility, which I didn't know about until recently. However, I'm puzzled why it's so hard to invoke, and am wondering if I'm missing something.

Consider this example:

#ifdef A
  printf("A\n");
#endif
#ifdef B
  printf("B\n");
#endif

If you invoke unifdef with "unifdef -DA junk.c", you get:

  printf("A\n");
#ifdef B
  printf("B\n");
#endif

Because you didn't tell unifdef that B was undefined, it didn't take it out.

I would like the utility to behave such that when I say unifdef -DA, I get instead:

  printf("A\n");

This would correspond to what the C preprocessor is actually doing: whatever branches are undefined are omitted.

To get this behavior with unifdef, I seem to need to use "unifdef -DA -UB junk.c", explicitly telling it that B is undefined. Though maybe I missed a simpler way to invoke it.

I wrote a python script to generate the long list of required -D and -U flags from the Makefile of the code I'm using (typically 80 per routine). And the results are excellent. But I'm wondering whether such a script is actually necessary.

It's also possible that another utility (sunifdef? coan?) has my desired behavior built in already; if so, please mention it.

Answer 1

The coan utility does what you need with the -m flag:

$ coan source -DA -m test.c 
  printf("A\n");

From the man page:

-m, --implicit
    Assume that any symbol that is not --define-ed is implicitly
    --undef-ed.