in reply to Exploiting Perls idea of what is a number

$temp1=eval { 0+$value1 };
$temp1=$value1 if $@; Perl says it is not a number

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 Ummm ... under what circumstances can that first line of code possibly set $@ ?

Update: moritz has provided a very plausible answer to that question.

Cheers,
Rob

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Re^2: Exploiting Perls idea of what is a number
by moritz (Cardinal) on Jul 08, 2008 at 13:46 UTC
    Under the influence of use warnings FATAL => qw(numeric);
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      Under the influence of use warnings FATAL => qw(numeric);

      Actually, just writing use warnings; is sufficient. Example: 

      use strict;
use warnings;
my $value="abc";
my $t=eval { 0+$value };
print "$@\n";

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      prints Argument "abc" isn't numeric in addition (+) at U:\develsv\ARTS\playground\eval_test1.pl line 4.

      -- 
      Ronald Fischer <ynnor@mm.st>
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        It does print that, but not because $@ is set. Here's a slight modification: 

        use strict;
use warnings;
my $value="abc";
my $t=eval { 0+$value };
print "OMG:  $@\n";
__END__
Argument "abc" isn't numeric in addition (+) ...
OMG:

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