R help - Mar 2005 - sub('^','var',1:3) produces unexpected results

Hi,

According to help(sub), the ^ should match the zero-length string at the
beginning of a string:

sub('^','var',1:3) # "1" "2" "3"
sub('$','var',1:3) # "1var" "2var"
"3var"

# This generates what I expected from the first case:
sub('^.','var',11:13)  # "var1" "var2"
"var3"

Dave
-- 
 Dr. David Forrest
 drf at vims.edu                                    (804)684-7900w
 drf5n at maplepark.com                             (804)642-0662h
                                   http://maplepark.com/~drf5n/

>>>>> "David" == David Forrest <drf5n at
maplepark.com>
>>>>>     on Tue, 22 Mar 2005 15:02:20 -0600 (CST) writes:
    David> According to help(sub), the ^ should match the
    David> zero-length string at the beginning of a string:

yes, indeed.

    David> sub('^','var',1:3) # "1" "2"
"3"
    David> sub('$','var',1:3) # "1var"
"2var" "3var"

    David> # This generates what I expected from the first case:
    David> sub('^.','var',11:13)  # "var1"
"var2" "var3"

there are even more fishy things here:

1) In your cases, the integer 'x' argument is auto-coerced to
   character, however that fails as soon as  'perl = TRUE' is used.

 > sub('^','v_', 1:3, perl=TRUE)
 Error in sub.perl(pattern, replacement, x, ignore.case) : 
	 invalid argument

 {one can argue that this is not a bug, since the help file asks
  for 'x' to be a character vector; OTOH, we have
  as.character(.) magic in many other places, i.e. quite
  naturally here;  
  at least  perl=TRUE and perl=FALSE should behave consistently.}


2) The 'perl=TRUE' case behaves even more problematically here:

  > sub('^','v_', LETTERS[1:3], perl=TRUE)
  [1] "A\0e" "B\0J" "C\0S"
  > sub('^','v_', LETTERS[1:3], perl=TRUE)
  [1] "A\0J" "B\0P" "C\0J"
  > sub('^','v_', LETTERS[1:3], perl=TRUE)
  [1] "A\0\0" "B\0\0" "C\0m" 
  >

 i.e., the result is random nonsense.

Note that this happens both for R-patched (2.0.1)  and R-devel (2.1.0 alpha).

==> "forwarded" as bug report to R-bugs