Text or Number

[Sisyphus]

Andrew Savige wrote:
> Joshua Goodall <joshua at roughtrade.net> wrote:
> 
>>#!/usr/bin/perl
>>use B qw(svref_2object);
>>sub cmpflags($$) { $_[1] & svref_2object($_[0])->FLAGS ? 1 : undef }
>>sub isinteger($) { cmpflags \$_[0], B::SVf_IOK | B::SVp_IOK }
>>sub isstring ($) { cmpflags \$_[0], B::SVf_POK | B::SVp_POK }
>>
>>$a=42; $b=$a; $c=$a; $d="$c";
>>for $i ($a, $b, $c, $d) {
>>    printf "%2s: int=%1s str=%1s\n", $i, isinteger($i), isstring($i);
>>}
> 
> 
> for $i ($a, $b, $c, $d) {
>     my $is_string = (~$i & $i) ne '0';
>     printf "%2s: int=%1s str=%1s\n", $i, !$is_string, $is_string;
> }
> 
> This old merlyn trick exploits that bitwise operators ~ and & operate
> differently on numbers and strings. Notice that Joshua's prints:
> 
> 42: int=1 str= 
> 42: int=1 str= 
> 42: int=1 str=1
> 42: int=  str=1
> 
> while merlyn's trick prints:
> 
> 42: int=1 str= 
> 42: int=1 str= 
> 42: int=1 str= 
> 42: int=  str=1
> 
> Why the difference?

The assignment $d = "$c" is altering the flags associated with $c - 
which produces that strange result in running Joshua's code.

If you take a Devel::Peek::Dump($c) immediately after the assignment 
$c=$a, you'll see FLAGS = (IOK,pIOK). If you take another 
Devel::Peek::Dump($c) after the assignment $d="$c", you'll see FLAGS = 
(IOK,POK,pIOK,pPOK).

Mind you, I don't know why that should be so :-)


 > Also, I'm curious to know which type of applications
> care about this (in the original question, I didn't understand exactly
> how this was causing trouble).
> 

Yes I'm not sure I understand, either - though my guess is that the OP 
would want *all* of the above to be reported as a number - which is what 
the perl API function  'looks_like_number()' would do, if I'm not mistaken.

Cheers,
Rob

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