On 5/24/05, Adrian Taylor <[EMAIL PROTECTED]> wrote:
> eval_is('undef + 1', undef, 'undef + 1', :todo<bug>); # dies
In this case, you're expecting
(undef) + 1
but you're getting
undef(+1)
instead.
This is because 'undef' serves double-duty as both 'undefined value'
and 'prefix op for undefining variables', and the op form is taking
precedence in parsing.
It's worth noting that Perl5 seems to have exactly the same parsing behaviour:
$ perl -e 'print undef + 1'
Warning: Use of "undef" without parentheses is ambiguous at -e line 1.
Can't modify constant item in undef operator at -e line 1, at EOF
Execution of -e aborted due to compilation errors.
I'm not sure whether this behaviour is supposed to be changing.
> eval_is('1 + undef', undef, '1 + undef', :todo<bug>); # gives 1
> eval_is('2 * undef', undef, '2 * undef', :todo<bug>); # gives 0
AFAIK, a vanilla (undef) numifies to 0, so these are already correct.
If you were expecting undef to behave as a 'poisoned' value then I can
understand the confusion.
> eval_is('undef xx 2', undef, 'undef xx 2', :todo<bug>); # dies
Again, you're expecting
(undef) xx 2
but you're getting
undef(xx 2)
instead.
Stuart
