[ILUG] Wildcards in makefile

[Brian Foster]

  | Date: Mon, 27 Nov 2006 16:34:13 +0000
  | From: Conor Daly <conor.daly_ilug at cod.homelinux.org>
  | 
  |[... since ] $(MANDIR)/man3/ will contain
  | manpages from other programs / libs, I cannot use:
  | 
  | 	rm -f $(MANDIR)/man3/*; \
  | 
  | for the uninstall.  Is there a form of 
  | 
  |     for FILE in ../man3/*; do rm -f $(MANDIR)/man3/$FILE; done
  | 
  | that I can use in a makefile to accomplish this?

 yes.  that, modified to be  (1) correct,
 and  (2) take into account make(1) syntax.

 the correctness issue is yer FILE variable
 is of the form ../man3/foo.3 so the rm(1)
 is removing  <mandir>/man3/../man3/foo.3
 which will usually, but not always, work.
 it may fail if  <mandir>/man3  is a symlink.

 the `make' syntax issue is to use a $ in
 a rule, it must be written $$.

 hence, something like either of the following
 should work (neither tested!):

     for FILE in ../man3/*; do \
        rm -f $(MANDIR)/man3/$$(basename "$$FILE"); \
     done
 or:
     for FILE in ../man3/*; do \
        rm -f $(MANDIR)/man3/$${FILE##*/}; \
     done

 if you are using GNU `make', there are various
 built-in macros which can help.  one possiblity
 might be (not tested!):

    $(foreach file,$(wildcard ../man3/*), \
        rm -f $(MANDIR)/man3/$(notdir $(file)))

 but, since that is harder to read, and I'm not too
 certain I got the syntax right, and it works only
 with GNU make (and maybe `clearmake'?), and has no
 (in this case) significant advantages, I'd use one
 of the first two, shell-embedded-in-Makefile, forms.

cheers!
	-blf-
-- 
Experienced (20+ yrs) kernel/software Eng: | Brian Foster   Montpellier,
 • Unix, embedded, &tc;  • Linux;  • doc;  | blf at utvinternet.ie   FRANCE
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