> -Original Message-
> From: Luke Palmer [mailto:[EMAIL PROTECTED]
> Sent: Tuesday, March 29, 2005 5:43 AM
> To: Zhuang Li
> Cc: Jeff Yoak; fwp@perl.org; perl6-language@perl.org
> Subject: Re: Unknown level of hash
>
> Zhuang Li writes:
> > Yes. I think it's both useful and fun. I was th
Zhuang Li writes:
> Yes. I think it's both useful and fun. I was thinking something similar
> to
> @[EMAIL PROTECTED] = map{1} @a;
>
> But getting "$hash->{E1}->{E2}->...->{En} = 1;" instead of "$hash{E1} =
> 1; ... $hash{En} =1;".
Yeah, like this:
%hash{dims @a} = (1) xx Inf;
> What I'd r
Zhuang Li wrote:
Yes. I think it's both useful and fun. I was thinking something similar
to
@[EMAIL PROTECTED] = map{1} @a;
But getting "$hash->{E1}->{E2}->...->{En} = 1;" instead of "$hash{E1} =
1; ... $hash{En} =1;".
What I'd really like to do is:
Given @a = ('E1', 'E2', ..., 'En');
@b
On Tue, Mar 29, 2005 at 06:42:41PM +0200, Alexandre Jousset wrote:
> Alexandre Jousset wrote:
> >Andrew Pimlott wrote:
> >
> (You should be able to write the first one as
>
> ${fold_left { \${$_[0]}->{$_[1]} } \$hash, @a} = 1;
>
> Is it a copy/paste or a retype ? Because there
Alexandre Jousset wrote:
Andrew Pimlott wrote:
(You should be able to write the first one as
${fold_left { \${$_[0]}->{$_[1]} } \$hash, @a} = 1;
Is it a copy/paste or a retype ? Because there's a comma missing after
the first closing brace...
but Perl complains for no reason I can see.)
--
Andrew Pimlott wrote:
(You should be able to write the first one as
${fold_left { \${$_[0]}->{$_[1]} } \$hash, @a} = 1;
but Perl complains for no reason I can see.)
Of course, the '{}' operator is a hash ref generator, not an
anonymous sub declarator as 'sub {}' is...
But I declared fold_left
On Tue, Mar 29, 2005 at 11:30:49AM +0200, Alexandre Jousset wrote:
> Andrew Pimlott wrote:
>
> > If you're feeling functional,
> >
> > ${fold_left(sub { \${$_[0]}->{$_[1]} }, \$hash, @a)} = 1;
> >
> > To get the value back out:
> >
> > fold_left { $_[0]->{$_[1]} } $hash, @a;
> >
> > Here i
Суханов Вадим wrote:
[snip code...]
gives:
$VAR1 = {
'E1' => {
'E2' => {
'E3' => {
'En' => 1
}
},
'E3' => {
В сообщении от 29 Март 2005 01:08 Alexandre Jousset написал(a):
> Hey all...
>
> Vladi Belperchinov-Shabanski wrote:
> [snip...]
>
> > and 'hidden loop' one:
> >
> >@a = ( 'E1', 'E2', 'E3', 'En' );
> >$a = 1;
> >map{ $a = { $_ => $a } } reverse @a;
> >print Dumper( $a );
>
> The p
Yitzchak Scott-Thoennes wrote:
I'm sorry but that doesn't work as is on my system: "Not a HASH
reference at x.pl line 9.".
Works for me. Can you show the actual code you are trying?
In fact I made a mistake. Sorry again, you're right ! :-)
--
\^/
-/ O \--
On Tue, Mar 29, 2005 at 11:45:01AM +0200, Alexandre Jousset wrote:
> Yitzchak Scott-Thoennes wrote:
> >To set it, you need to just do the same thing except that at each step
> >instead of keeping the hash element, you keep a reference to it:
> >
> >$entryref = \$hash;
> >$entryref = \$$entryref->{$
Yitzchak Scott-Thoennes wrote:
To set it, you need to just do the same thing except that at each step
instead of keeping the hash element, you keep a reference to it:
$entryref = \$hash;
$entryref = \$$entryref->{$_} for @a;
$$entryref = 1;
Note that this will create hashrefs at any undefined level
Hello,
Andrew Pimlott wrote:
> If you're feeling functional,
>
> ${fold_left(sub { \${$_[0]}->{$_[1]} }, \$hash, @a)} = 1;
>
> To get the value back out:
>
> fold_left { $_[0]->{$_[1]} } $hash, @a;
>
> Here is fold_left:
>
> sub fold_left (&@) {
> my $sub = shift;
>
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