Mistakenly Believing split is using @

enoch has asked for the wisdom of the Perl Monks concerning the following question:

Perl gives the warning: Use of implicit split to @_ is deprecated when used in the following code:

print scalar((split('', "I am a string"))) . "\n";
[download]

Obviously, I am not making use of @_ in this split. Is perl warning because I am not assigning the return value of the split to an array? Because if I do this:

my @chars = split '', "I am a string";
my $num = scalar(@chars);
print $num . "\n";
[download]

It does not warn. I am not really worried about the warning because the output ('13' in this case) is correct. I was just wondering if this is a bug, or if this is by design. And, if it is by design, why? Or, perhaps somewhere in the parsing, perl becomes confused with the scalar call on the returned array of split. I don't know. I am just wondering. I have tried this on both perl v5.6.1 for i686-linux and ActiveState perl v5.6.0 binary build 623 and received the same warning.

Jeremy

Comment on Mistakenly Believing split is using @_ Select or Download Code

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Re: Mistakenly Believing split is using @_ by Maestro_007 (Hermit) on Sep 19, 2001 at 22:15 UTC
It's doesn't warn as a result of what you're splitting on, but what you're assigning to. In the olden days, the idiom would look like this: `split;` [download] which would automatically assign the contents to `@_`. Maybe the warning should say `Implicitly assigning result of split() to @_ is deprecated`. I got burned on that one once upon a time, myself. The reason the second snippet doesn't warn is that you have assigned the result to `@chars`. I think it's safe to say that the newer behavior is a Good Thing. As to why the warning doesn't understand that you are still using the results in another context is a little puzzling, something that merlyn can probably answer. MM Update: void context =~ scalar context, got it. Thanks japhy!	[reply] [d/l] [select]
Re: Mistakenly Believing split is using @_ by lemming (Priest) on Sep 19, 2001 at 22:22 UTC
From perldoc perlfunc: In scalar context, returns the number of fields found and splits into the "@_" array. Use of split in scalar context is deprecated, however, because it clobbers your subroutine arguments. It's preventing you from possibly shooting yourself in the foot. For a better method, use the function length() to count the number of characters in a string.	[reply]
Re: Mistakenly Believing split is using @_ by japhy (Canon) on Sep 19, 2001 at 22:15 UTC
`split()` in scalar context (update: void context is a special type of scalar context) assigns to `@_`. That's documented. Chances are you can use something other than `split()` if you just want a field-count. _____________________________________________________ Jeff`[japhy]`Pinyan: Perl, regex, and perl hacker. `s++=END;++y(;-P)}y js++=;shajsj<++y(p-q)}?print:??;`	[reply]
Re: Mistakenly Believing split is using @_ by Anonymous Monk on Sep 19, 2001 at 22:18 UTC
If you are trying to get the number of characters in a string you should use length instead of split('').	[reply]
Re: Mistakenly Believing split is using @_ by particle (Vicar) on Sep 19, 2001 at 22:20 UTC
i believe you are using the depricated feature. if you don't set the value of split to an array, such as `my @chars=split '', "I am a string";`, then it's set to @_. `{ print scalar(my @chars=(split('', "I am a string"))) . "\n"; }` [download] works fine. the scoping should eliminate the @chars var when you're done with it. although, you might try `length` as in `print length("I am a string"), "\n";` [download] ~Particle	[reply] [d/l] [select]


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Mistakenly Believing split is using @_