The pre-increment and post-increment operators are designed to work on strings too. Witness the following:

my $string = "abc";
print ++$string, "\n";
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If my calculations are correct, you should see "abd".

So the point is that if you give ++ a string that it can figure out, it will increment it. Thus, non-numeric strings are legal arguments for ++. Now, in your case, you gave the increment operator a string that it couldn't sensibly increment in its non-numeric form, so the ++ operator dropped into numeric form, and incremented 0 to 1. No warning because non-numeric strings are legal in this case.

To read more about the ++ operator's "magic", see perlop.

I see two holes in your argument.

1) That particular string is non-numeric and it does get converted to numeric. What would be the point of not issuing a "non-numeric isn't numeric" warning when it happens because the operator works with some other strings?

2) perl -wle "$s='abc'; $s--; print $s" doesn't give a warning either, yet non-numeric strings are never legal for the post-decrement operator.

I suspect the warning is surpressed as a side-effect (intentional or not) of supressing the undefined warning when executing my $i; $i++;.

However, the ++'s operator's magic applies to strings that =~ m/^[a-zA-Z]*[0-9]*\z/. "a,b" doesn't match.

I suspect that in the third case perl is calling int($m), which returns 0, then applying the prefix increment operator.

Code tags added by GrandFather

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non-numeric strings are legal arguments for ++

++ works even on undefined values:

$ /usr/bin/perl -le'print $a++'
0
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--
David Serrano

$m="acb";
print ++$m; # Prints acc, as expected (according to perlop)
# and throws no warning ... wierd huh?
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