On 08/02/2015 02:35 PM, Lloyd Fournier wrote:
> @a = $z[0].list
>
> or in a less documented way:
>
> @a = $z[0]<>
>
> The idea is that you can store an array in ether a @ or $ variable.
> Where as in perl5 you could only store a reference in $ variable. The
> sigil simply tells perl6 how it shoul
On Sun, Aug 02, 2015 at 10:35:23PM +1000, Lloyd Fournier wrote:
> If you want to assign to an array which is an element of another array:
>
> @a = $z[0].list
this is very confusing because this is not a LoL. $z.flat seems more
intuitive to me.
--
Marc Chantreux (eiro on github and freenode)
htt
Having followed Perl6 from its inception, it is good to see Christmas
will be coming this year :)
Richard
On 08/03/2015 08:49 AM, Larry Wall wrote:
On Sun, Aug 02, 2015 at 03:09:07PM +0200, Moritz Lenz wrote:
: If you want to extract it from the scalar, use the @ again, this
: time as a prefi
Non expert opinion here.
my $z = ['a', 'b', 'c'];
Is a real array. The difference is that is in a non-auto flattening
container (variable). If you want to iterate over it you can:
for @$z { ... } # like perl5
If you want to assign it to @ array then do the same thing @a = @$z.
If you want to a
On Mon, Aug 3, 2015 at 3:49 AM, Larry Wall wrote:
> On Sun, Aug 02, 2015 at 03:09:07PM +0200, Moritz Lenz wrote:
> : If you want to extract it from the scalar, use the @ again, this
> : time as a prefix:
> :
> : for @$s { } # two iterations again.
>
> Note that this distinction will go away after
On Sun, Aug 02, 2015 at 03:09:07PM +0200, Moritz Lenz wrote:
: If you want to extract it from the scalar, use the @ again, this
: time as a prefix:
:
: for @$s { } # two iterations again.
Note that this distinction will go away after the Great List Refactor,
so Gabor gets his wish. Part of the G
> It's an Array inside a Scalar.
This is really disturbing from a newcommer perspective. I don't know
much about the Perl6 internals so maybe i can explain it better (or
someone can fix me):
Let's try:
Everything is an object in Perl6, which comes with roles and
inheritances and [Sigils](http:/
This whole thing sounds quite complex. What is the value behind this
complexity?
Gabor
Hi,
On 02.08.2015 06:43, Gabor Szabo wrote:
On Fri, Jul 31, 2015 at 4:16 PM, Moritz Lenz mailto:mor...@faui2k3.org>> wrote:
On 07/31/2015 03:02 PM, Gabor Szabo wrote:
The following code (with comments) is confusing me.
Can someone give some explanation please?
S
On Fri, Jul 31, 2015 at 4:16 PM, Moritz Lenz wrote:
>
>
> On 07/31/2015 03:02 PM, Gabor Szabo wrote:
>
>> The following code (with comments) is confusing me.
>> Can someone give some explanation please?
>> Specifically the difference between
>>
>> my @x = ;
>>
>
> It's the assignment to the Array
On 07/31/2015 03:02 PM, Gabor Szabo wrote:
The following code (with comments) is confusing me.
Can someone give some explanation please?
Specifically the difference between
my @x = ;
It's the assignment to the Array variable that makes the Array here; < >
by itself just creates a Parcel:
The following code (with comments) is confusing me.
Can someone give some explanation please?
Specifically the difference between
my @x = ;
my $z = @x;
and
my $z = ;
Gabor
use v6;
# < > creates an array here:
my @x = ;
say @x.WHICH; # Array|140713422866208
say @x;# a b
say @x[0];
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