Sean O'Rourke skribis 2004-04-19 15:34 (-0700): > I'm saying "division" is now defined such that when the numerator is > a hash(-ref), the result is the set of values associated with the > denominator. I've never tried to divide a hash or hashref by > something without it being a bug.
I understand now. But that means the meaning of the / is unknown until runtime, which means $foo/0 can't be a compile time error. And it doesn't quote the thing after it, which means still doing a lot of typing. $foo/bar should be a compile time error (Perl 6 has no barewords) if $foo is not a hashref, but be $foo{'bar'} if it is. Waiting for runtime is bad, I think. > /foo/ # trailing slash -- so it's a regexp (m/foo/) > /foo\/bar/ # trailing slash -- syntax error (m/foo/ bar/) > /foo/a # hash-path -- no trailing slash ($_.{'foo'}{'a'}) > /foo\/bar # hash-path -- no trailing slash ($_.{'foo/bar'}) > /foo\/ # hash-path -- no trailing slash ($_.{'foo/'}) Thanks. Now I'm sure I don't like the bare path idea. After a hash, perhaps it's doable, and even "if -r /etc/passwd" is doable, but there are too many allowed characters in filenames (on my system: any character except \0 and /). Juerd