%% Lin George <[EMAIL PROTECTED]> writes: >> foo:: bar ; echo Update due to bar
>> foo:: baz ; echo Update due to baz lg> Good example! I think the effect is the same as the following lg> statements (I think either bar or baz is out of date, foo will be lg> remade.), lg> foo: bar baz ; echo something But this doesn't do the same thing. In my example above, "Update due to bar" will be printed if foo is out of date with respect to bar, and "Update due to baz" will be printed if foo is out of date with respect to baz. If neither is out of date, nothing is printed. If both are out of date, both are printed. In your example, "something" is printed if either or both of bar and baz are out of date. lg> except that it is not easy to detect whether foo is remade by bar lg> or remade by baz. Is that correct? No, it's easy to tell which prerequisites were out of date. Look up the $? variable in the GNU make manual. -- ------------------------------------------------------------------------------- Paul D. Smith <[EMAIL PROTECTED]> Find some GNU make tips at: http://www.gnu.org http://make.paulandlesley.org "Please remain calm...I may be mad, but I am a professional." --Mad Scientist _______________________________________________ Help-make mailing list [email protected] http://lists.gnu.org/mailman/listinfo/help-make
