Re: perl 'advancing' variable facility..
- From: paul@xxxxxxxx (Paul Johnson)
- Date: Wed, 25 May 2005 12:39:55 +0200
On Wed, May 25, 2005 at 10:27:11AM +0000, Vineet Pande wrote:
> Hi:
> I don't understand the way perl 'advances' the variables along the range
> a-z, A-Z, and 0-9.
> For example from a book
>
> $a = "A9"; print ++$a, "\n";
>
> gives B0 as expected,
>
> but,
> $a = "Zz"; print ++$a, "\n";
>
> gives AAa.
>
> Why?? why not Aa only!
Because you get a carry, similar to the way 99 + 1 = 100, not 00.
> also $a = "9z"; print ++$a, "\n";
>
> gives 10.
> and why not 0a?
Because "9z" doesn't match the criteria for magic auto-increment.
perldoc perlop says:
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'
--
Paul Johnson - paul@xxxxxxxx
http://www.pjcj.net
.
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