On Sun, 2010-02-07 at 06:24 -0800, Mark Galeck (CW) wrote: > OK, let me make sure I am not confused: > > Are you saying that on Unix: > > >echo \\\foobar > \foobar > > And that is why you expected that?
Yes. Well, to be precise I expected it because I know how backslash escaping works in the shell. Without any quoting, the first backslash escapes the second backslash, giving a single backslash, and the third backslash escapes the "f". Since an escaped character is just that character, "\f" is just "f", and the expected result is "\foobar". -- ------------------------------------------------------------------------------- Paul D. Smith <[email protected]> Find some GNU make tips at: http://www.gnu.org http://make.mad-scientist.net "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
