On Sun, May 13, 2001 at 11:37:01PM -0500, Me wrote:
Yes. But I'm not sure that:
# ordered
@array = (1, 2, 3, 5, 8);
# unordered
%hash = (Fred = 22, Jane = 30);
is more or less typical than:
# unordered:
@array = ('England', 'France', 'Germany');
# ordered:
On Thu, 10 May 2001 17:15:09 +0100, Simon Cozens wrote:
What you could do, is treat an iterator as something similar to reading
a line from a file. Tied filehandles allow something like it in Perl5.
You know, if what you say is true, I'd expect to find a module on CPAN which
turns the exotic
On Mon, May 14, 2001 at 01:25:51PM +0200, Bart Lateur wrote:
There must be some reason why a language like Sather isn't more popular.
I think that iters are part of the problem.
That smacks of the Politician's Syllogism:
Something is wrong.
This is something.
Therefore this is
an ordered hash is common
Arrays too.
not wise ... to alter features just for beginners.
Agreed.
(PS 11 people isn't a statistic, its a night at the pub)
Your round...
The extra complexity of a separate hash syntax might
be justified for other reasons, but not the ones
On Mon, May 14, 2001 at 12:32:37PM -0500, Me wrote:
an ordered hash is common
Arrays too.
not wise ... to alter features just for beginners.
Agreed.
(PS 11 people isn't a statistic, its a night at the pub)
Your round...
The extra complexity of a separate
Hm, OK. What does this access and using what method ?
$foo = '1.2';
@bar[$foo];
This is an argument against conflating @ and %.
It has nothing to do with using [] instead of {}.
(I accept that the @/% issue is problematic. Otoh,
I don't yet see @/% conflation as being obviously
a bad
On Mon, May 14, 2001 at 01:56:01PM -0500, Me wrote:
Hm, OK. What does this access and using what method ?
$foo = '1.2';
@bar[$foo];
This is an argument against conflating @ and %.
No it is not.
It has nothing to do with using [] instead of {}.
Yes it does. I was asking if the
On Mon, May 14, 2001 at 03:23:56PM -0400, Buddha Buck wrote:
At 08:10 PM 05-14-2001 +0100, Graham Barr wrote:
On Mon, May 14, 2001 at 01:56:01PM -0500, Me wrote:
Hm, OK. What does this access and using what method ?
$foo = '1.2';
@bar[$foo];
This is an argument against
Damian Conway wrote [and John Porter reformats]:
@bar[$foo]; # Access element int($foo) of array @bar
%bar{$foo}; # Access entry $foo of hash %bar
@bar{$foo}; # Syntax error
%bar[$foo]; # Syntax error
And why is that superior to:
@bar[$foo]; # Access element int($foo) of array @bar
When all the smoke clears, it will be relatively simple to declare an
ordered hash probably on the order of adding a single word to its
declaration.
Yep. In fact, it's now relatively simple in Perl 5.
You just grab the Attribute::Handlers and Tie::SortHash modules and
add a single
On Mon, May 14, 2001 at 03:41:24PM -0400, John Porter wrote:
Damian Conway wrote [and John Porter reformats]:
@bar[$foo]; # Access element int($foo) of array @bar
%bar{$foo}; # Access entry $foo of hash %bar
@bar{$foo}; # Syntax error
%bar[$foo]; # Syntax error
And why is that
@bar[$foo]; # A
%bar{$foo}; # B
@bar{$foo}; # C
%bar[$foo]; # D
You forgot
$bar[$foo]; # $bar is an array reference
$bar{$foo}; # $bar is a hash reference
I can't argue with that.
My vote is now against conflating [] and {}.
Graham Barr wrote:
As I said in another mail, consider
$bar[$foo];
$bar{$foo};
But if @bar is known to be one kind of array or
the other, where is the ambiguosity that that is
meant to avoid?
--
John Porter
On Mon, May 14, 2001 at 03:58:31PM -0400, John Porter wrote:
Graham Barr wrote:
As I said in another mail, consider
$bar[$foo];
$bar{$foo};
But if @bar is known to be one kind of array or
the other, where is the ambiguosity that that is
meant to avoid?
I did not say it was
On Mon, May 14, 2001 at 02:51:08PM -0500, Me wrote:
survey ? I never saw any survey,
It was an informal finger-in-the-wind thing I sent to
a perl beginners list. Nothing special, just a quick
survey.
http://www.self-reference.com/cgi-bin/perl6plurals.pl
As someone else pointed out (I
On Mon, 14 May 2001 20:38:31 +0100, Graham Barr wrote:
You forgot
$bar[$foo]; # $bar is an array reference
$bar{$foo}; # $bar is a hash reference
As to what the combined
$bar[$foo]
would mean: that depends on what $bar contains.
(Aw! That hurt!)
--
Bart.
Bart Lateur wrote:
As to what the combined
$bar[$foo]
would mean: that depends on what $bar contains.
I think it would depend on what the declared type
of @bar was (i.e. ordered or associative).
--
John Porter
As someone else pointed out (I forget who). But beginners are not
always the best people to ask. Beginner don't stay beginners for
long I think the quote was.
And as I said before, I agree.
I picked the beginners list as much because it was
active as anything else. They are *somebody* after
On Mon, May 14, 2001 at 10:11:01PM +0200, Bart Lateur wrote:
On Mon, 14 May 2001 20:38:31 +0100, Graham Barr wrote:
You forgot
$bar[$foo]; # $bar is an array reference
$bar{$foo}; # $bar is a hash reference
As to what the combined
$bar[$foo]
would mean:
(i.e. ordered or associative).
A (probably futile, but one has to try) plea for
people to use numbered rather than ordered.
@foo = ['England', 'France', 'Germany'];# unordered
%foo = {First = Fred', Last = 'Bloggs']; # ordered
(I'd also suggest named instead of the scientific
On Mon, May 14, 2001 at 08:38:31PM +0100, Graham Barr wrote:
What is the meaning of the following four expressions in Perl6?
@bar[$foo];
%bar{$foo};
@bar{$foo};
%bar[$foo];
$bar[$foo];
$bar{$foo};
It's really, really easy. Just
Edward Peschko wrote:
As to what the combined
$bar[$foo]
would mean: that depends on what $bar contains.
I like visual clues to tell me
what type of variable
something is. And I disagree strongly with trying to
steamroller the language's
design paper-flat as much as I
On Mon, May 14, 2001 at 01:25:51PM +0200, Bart Lateur wrote:
There must be some reason why a language like Sather isn't more popular.
I think that iters are part of the problem.
That smacks of the Politician's Syllogism:
Something is wrong.
This is something.
Therefore this
Simon Cozens wrote:
It's really, really easy. Just stick a - between the variable and
the brace, and you have Perl 5.
Pardon my indelicacy, but - Screw how it looks in Perl5.
We can make it mean anything, and appear anyhow we want.
IOW, what makes sense in Perl6 isn't defined by how
we were
On Mon, May 14, 2001 at 04:50:17PM -0400, John Porter wrote:
Pardon my indelicacy, but - Screw how it looks in Perl5.
I'm not telling you how it *looks* in Perl 5, I'm telling you (in Perl 5
terms) what it will *mean*.
--
Use an accordion. Go to jail.
-- KFOG, San Francisco
Simon Cozens wrote:
I'm not telling you how it *looks* in Perl 5, I'm telling you (in Perl 5
terms) what it will *mean*.
Fine, you're using perl5 as pseudocode.
I could do that too. But it has no bearing on the
desirability of anyone's proposed perl6 syntax or
semantics.
--
John Porter
On Mon, May 14, 2001 at 04:50:17PM -0400, John Porter wrote:
Pardon my indelicacy, but - Screw how it looks in Perl5.
I'm not telling you how it *looks* in Perl 5, I'm telling you (in Perl 5
terms) what it will *mean*.
nice save
p
I will read replies, and respond to off list emails,
but I will refrain from posting to the list on this topic
for at least one week.
If you have nothing new to add, then please don't post.
Suggestion: pseudohash.
%foo{Fred} = 'Bloggs';
$bar = %bar[1];# $bar is 'Bloggs'
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