> Bryan R Harris wrote:
>>
>>>> 2) perl -le '$x = qw/a b c d e/; print $x'
>>>> e
>>>> Or better what is (2) doing?
>>> Read up on the comma operator in perlop (and I know there are no literal
>>> commas in (2) but qw/a b c d e/ behaves exactly the same as ('a', 'b', 'c',
>>> 'd', 'e').)
>>>
>>> perldoc perlop
>>
>>
>> Why would this be considered a binary (?) use? I read the documentation,
>> but I think I'm missing something, it doesn't make sense.
>>
>> This is really bizarre and unexpected for me (I'm not the OP, by the way).
>> Can someone help train my intuition as to why these aren't identical? And
>> why they shouldn't be?
>>
>> % perl -le '$x = @{[ "a","b","c","d","e" ]}; print $x'
>> 5
>>
>> % perl -le '$x = ("a","b","c","d","e"); print $x'
>> e
>
> @{} is an array so it returns the number of elements. (The anonymous array in
> [] is dereferenced by @{}.)
>
> "a","b","c","d","e" is a list so each element is evaluated and discarded and
> the last element is assigned to the scalar.
Ah. What I was missing is that a list is *not* the same as an array.
Interesting, thanks.
For those following this thread (probably nobody), this is demonstrated
exactly in Programming Perl, 24.1.2 (I just found it right now).
- B
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