Re: Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
On Nov 6, Dan Anderson said:
>
>> Dan> So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
>>
>> Only in recent Perls.
>
>Do you know exactly how recent? Are we talking 5 or better or 3 or
>better?
Without check perldeltas, I'd say 5.6.
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Re: Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
> Dan> So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
>
> Only in recent Perls.
Do you know exactly how recent? Are we talking 5 or better or 3 or
better?
Thanks in advance,
-Dan
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Re: Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
> "Dan" == Dan Anderson <[EMAIL PROTECTED]> writes:
Dan> So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
Only in recent Perls. The mapping of qw(...) to a (...) list at compile
time is a modern addition. Older Perls replaced it with a runtime
split on the string, and probably would not accept it as the arglist
of a method call.
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Re: Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
[sorry about that first post, I got ^X-happy]
On Nov 5, Dan Anderson said:
>use Data::Dump qw(dump);
>foo->bar qw(foo bar);
>Am I correct in assuming that if I have a subroutine foo (or method if
>called with a package name), and I use qw() it takes all words seperated
>by spaces, and passes them in as arguments.
>
>So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
The qw() operator changes your source code at compile-time, which is why
you can say
$object->method qw(...)
when ordinarily you'd need
$object->method(...)
When you use qw(this that those), Perl changes that to
('this', 'that', 'those')
Perl splits the qw(...) on spaces, and returns the raw data,
single-quoted. This means no variables. You can't even escape a space:
qw( abc\ def )
becomes
('abc\\', 'def')
That is all.
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Re: Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
On Nov 5, Dan Anderson said:
>I've noticed that in code examples something like the following will be
>used:
>
>use Data::Dump qw(dump);
>foo->bar qw(foo bar);
>
>(Syntax may not be 100% correct).
>
>Am I correct in assuming that if I have a subroutine foo (or method if
>called with a package name), and I use qw() it takes all words seperated
>by spaces, and passes them in as arguments.
>
>So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
>
>Thanks in advance,
>
>Dan
>
>
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Re: Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
Dan Anderson wrote:
> I've noticed that in code examples something like the following will be
> used:
>
> use Data::Dump qw(dump);
> foo->bar qw(foo bar);
>
> (Syntax may not be 100% correct).
>
> Am I correct in assuming that if I have a subroutine foo (or method if
> called with a package name), and I use qw() it takes all words seperated
> by spaces, and passes them in as arguments.
>
> So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
almost:
foo->bar ('foo', 'bar')
qq (foo bar) would render
("foo", "bar")
It does make a difference.
Joseph
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Is foo qw (arg1 arg2) equivalent to foo (arg1, arg2)?
I've noticed that in code examples something like the following will be
used:
use Data::Dump qw(dump);
foo->bar qw(foo bar);
(Syntax may not be 100% correct).
Am I correct in assuming that if I have a subroutine foo (or method if
called with a package name), and I use qw() it takes all words seperated
by spaces, and passes them in as arguments.
So: foo->bar qw(foo bar); is equivalent to foo->bar("foo","bar"); ?
Thanks in advance,
Dan
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