On 1/4/07, David Boyce <[EMAIL PROTECTED]> wrote: ...
The problem is that occasionally a target will need one of these removed as a special case. But of course using $(filter-out -z foo,...) will remove *all* the -z flags, leaving the other keywords hanging there alone, while quoting the string like $(filter-out "-z foo",...) doesn't match anything. Does anybody know a reasonable way to remove a whitespace-containing string?
Option 1) don't put a space between the -z and its argument to begin with, then use $(filter-out) Option 2) don't put the unwanted arguments in the general variable but rather keep them in another variable for use with just those targets that need them Option 3) $(subst -z foo,,$(args)) ...but that fails if there are multiple spaces or a tab between the -z and 'foo' Option 4) $(subst -z foo,,$(strip $(args)) ...but that replaces *all* internal whitespace in $(args) with single spaces Philip Guenther _______________________________________________ Help-make mailing list [email protected] http://lists.gnu.org/mailman/listinfo/help-make
