Resending due to BT doing bad things to good nameservers.
Damian Conway <[EMAIL PROTECTED]> writes:
>> >$val = (foo())[0];
>> >
>> > List?
>>
>> Scalar, obviously.
>
> How do you figure that? (Not a criticism: I'd really like to understand your
> thought process her
Damian Conway <[EMAIL PROTECTED]> writes:
>> >$val = (foo())[0];
>> >
>> > List?
>>
>> Scalar, obviously.
>
> How do you figure that? (Not a criticism: I'd really like to understand your
> thought process here so I can assess the relative DWIMity of the two
> alternat
Aaron Sherman wrote:
> >> $ref = [1,2];
> >> @ary[$ref] = foo(); # probably a syntax error
>
> Ok, as far as I can recall, Larry hinted that arrays and references to
> arrays would be interchangable in many contexts in P6. In this case, I
> can't see any reason that subscripting wou
> > $val = (foo())[0];
> >
> > List?
>
> Scalar, obviously.
How do you figure that? (Not a criticism: I'd really like to understand your
thought process here so I can assess the relative DWIMity of the two
alternatives).
> With a possible runtime error if foo doesn't ret
Aaron Sherman <[EMAIL PROTECTED]> writes:
> On Sun, Dec 16, 2001 at 03:55:10PM +1100, Damian Conway wrote:
>
> [...]
>
>>> And, just for laughs:
>>>
>>> $ref = [1,2];
>>> @ary[$ref] = foo(); # probably a syntax error
>
> Ok, as far as I can recall, Larry hinted that arrays a
On Sun, Dec 16, 2001 at 03:55:10PM +1100, Damian Conway wrote:
[...]
>> And, just for laughs:
>>
>> $ref = [1,2];
>> @ary[$ref] = foo(); # probably a syntax error
Ok, as far as I can recall, Larry hinted that arrays and references to
arrays would be interchangable in many
Piers posed the following puzzles:
> @ary[0] = foo() # scalar
Yes.
> @ary[1,2] = foo() # list context
Yes.
> @bar = 1;
> @ary[@bar] = foo() # ? probably list or maybe scalar...
List. With an explicit array as index, it's definitely a (one-element) slice.
> @bar = (1,
"Brent Dax" <[EMAIL PROTECTED]> writes:
> Piers Cawley:
> # In the following code fragment, what context is foo() in?
> #
> # @ary[0] = foo()
>
> The short answer is scalar context. The long answer is below. Note
> that the long answer is only the way I think of it. You may think
> differe
> > @ary[0] =()= foo(); # " " " " " "" "
>
> Hm, thats a change from perl5. In perl5 that would assign the number of
> elements returned from foo(). Is there a good reason for this change ?
Firstly, Larry may have to rule on which behaviour actually *is*
On Fri, Dec 14, 2001 at 06:39:02AM +1100, Damian Conway wrote:
>
>> In the following code fragment, what context is foo() in?
>>
>> @ary[0] = foo()
>
> Scalar context. @ary[0] is a single element of @ary.
>
> To call foo() in list context use any of the following:
>
> (@
On Thu, 13 Dec 2001 12:17:44 -0500, Mark J. Reed wrote:
> @i = (0);
> @ary[@i] = foo();
>
>How could one get that behavior without the intermediate array?
Parens, likely.
(@ary[0]) = foo();
--
Bart.
> In the following code fragment, what context is foo() in?
>
> @ary[0] = foo()
Scalar context. @ary[0] is a single element of @ary.
To call foo() in list context use any of the following:
(@ary[0]) = foo(); # Assign @ary[0] the first element returned
@(@ary[
Piers Cawley:
# In the following code fragment, what context is foo() in?
#
# @ary[0] = foo()
The short answer is scalar context. The long answer is below. Note
that the long answer is only the way I think of it. You may think
differently.
I like to think of it as 'one context'. 'Scalar'
On Thu, Dec 13, 2001 at 12:12:14PM -0500, Ted Ashton wrote:
> Thus it was written in the epistle of Piers Cawley,
> > In the following code fragment, what context is foo() in?
> >
> > @ary[0] = foo()
>
> Scalar, I would think.
I assume that the following would make the assignment a slice
an
Thus it was written in the epistle of Piers Cawley,
> In the following code fragment, what context is foo() in?
>
> @ary[0] = foo()
Scalar, I would think.
Just my guess,
Ted
--
Ted Ashton ([EMAIL PROTECTED]) | From the Tom Swifty collection:
Southern Adventist University| "Multiplicati
In the following code fragment, what context is foo() in?
@ary[0] = foo()
the following code
@ary= foo()
obviously evaluates @foo in a list context, but in the first I'm no
longer sure.
--
Piers
"It is a truth universally acknowledged that a language in
possession of a ri
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