8/13    Is there a unix command to print the full filesystem path of
        something? e.g.
        % pwd
        /home/troll
        % print-full-path drivel
        /home/troll/drivel
        \_ function print-full-path { (cd "$1"; pwd); }
           Implementation as a csh alias left as an exercise.
           \_ that doesn't work... $1 could be a file
        \_ You need something like:
           FDIR=`dirname "$relpath"`
           FBASE=`basename "$relpath"`
           FABS="`cd \"$D\" 2>/dev/null && pwd || echo \"$D\"`/$B"
           \_ christ. ok the answer is "no, you need a script".
        \_ which foo
        \_ Ok here's my little perl script for this, if anyone cares. -op
           \_ locate foo
              whereis foo
              find `pwd` -name foo -maxdepth 1
              echo `pwd`/foo
              \_ that last thing would work but I wanted symbolic links
              \_ `pwd`/foo would work but I wanted symbolic links
                 and relative paths converted to absolute paths.
                 The others don't work in general.
        \_ Here's my perl script for this, if anyone cares. -op
                #!/usr/bin/perl
                chop($cwd = `pwd`);
                foreach $arg (@ARGV) {
                  chdir $cwd;
                  chop($rel = `dirname $arg`);
                  chop($base = `basename $arg`);
                  chdir $rel || die "path not found: $rel\n";
                  chop($full = `pwd`);
                  print $full;
                  print "/" if ($full ne "/");
                  print $base if ($base ne "/");
                  print "\n";
                }
                \_ Look up chomp() as a chop() replacement for removing \n.