Amusement in Auto-Increments

Perl has an auto-increment function, and it's smart enough to recognise a string and increment within case.... but sometimes it's just too smart:

@foo = ("1zz", "zzz", "zz1", "z1z", "zz9");

foreach my $foo (@foo) {
  print "$foo autoincraments to ";
  $foo++;
  print "$foo\n";
}
[download]

gives

1zz autoincraments to 2
zzz autoincraments to aaaa
zz1 autoincraments to zz2
z1z autoincraments to 1
zz9 autoincraments to aaa0
[download]

Now, the first is easy, and the second is cool.... but the rest? Bizzare!

-- Ian Stuart
A man depriving some poor village, somewhere, of a first-class idiot.

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Re: Amusement in Auto-Increments by JavaFan (Canon) on Jul 22, 2009 at 11:09 UTC
Bizzare! Bizarre, but documented: The auto-increment operator has a little extra builtin magic to it. If you increment a variable that is numeric, or that has ever been used in a numeric context, you get a normal increment. If, however, the variable has been used in only string contexts since it was set, and has a value that is not the empty string and matches the pattern `"/^[a-zA-Z][0-9]\z/"`, the increment is done as a string, preserving each character within its range, with carry: print ++($foo = '99'); # prints '100' print ++($foo = 'a0'); # prints 'a1' print ++($foo = 'Az'); # prints 'Ba' print ++($foo = 'zz'); # prints 'aaa'	[reply] [d/l]
Re^2: Amusement in Auto-Increments by cog (Parson) on Jul 22, 2009 at 20:52 UTC
Using this bit of documentation, here's what happens: 1zz 1zz doesn't match `"/^[a-zA-Z][0-9]\z/"` Rule #1 - gets converted to a number, 1, which is then incremented to 2. zzz zzz matches `"/^[a-zA-Z][0-9]\z/"` Rule #2 - the last z is incremented to a (stays in the same range, which is a-z), carries one over; second z is incremented to a, carries one over; first z is incremented to a, carries one over, which gets turned into an a. zz1 zz1 matches `"/^[a-zA-Z][0-9]\z/"` Rule #2 - 1 is incremented to 2. z3z z3z doesn't match `"/^[a-zA-Z][0-9]\z/"` Rule #1 - the value is converted to numeric; since there are no digits before the first non-digit, the numeric value is 0; that value is then incremented to 1. zz9 zz9 matches `"/^[a-zA-Z][0-9]\z/"` Rule #2 - 9 is incremented to 0, carries one over, and then the process with the two zs is the same as from above.	[reply] [d/l] [select]
Re: Amusement in Auto-Increments by moritz (Cardinal) on Jul 22, 2009 at 10:49 UTC
FWIW this is what Rakudo currently does, and I believe it is correct: `1zz -> 2aa zzz -> aaaa zz1 -> zz2 z1z -> z2a zz9 -> aaa0 Z9Z -> AA0A` [download] It always starts at the last character, auto-increments that, and if there's an overflow it auto-increments the character to the left, recursively. It has the additional magic of stopping at a dot, so `img001.jpg` auto-incremented becomes `img002.jpg`.	[reply] [d/l] [select]
Re: Amusement in Auto-Increments by wol (Hermit) on Jul 28, 2009 at 11:36 UTC
`$foo = "Perl5"; $foo++; print $foo; ^Z Perl6` [download] Phew. Just had to check. -- use JAPH; print JAPH::asString();	[reply] [d/l]