Re: Which brackets should @a.perl use?
>> "m" == moritz writes:
>
> m> S02 says:
>
> m> "To get a Perlish representation of any object, use the .perl method.
> Like
> m> the Data::Dumper module in Perl 5, the .perl method will put quotes
> around
> m> strings, square brackets around list values,"
>
> m> So according to this, Rakudo has it right.
> m> But I think that a .perl()ification as ("blue", "light", "hayard",)
> would
> m> make much more sense, because simple thing like
>
> m> @a.push eval(@b.perl)
>
> m> would then DWIM.
>
> for your def of DWIM. i can see wanting an anon array to be pushed onto
> @a building up a structure.
DWIM in the sense that "eval($stuff.perl)" should behave the same as
$stuff itself.
since @a.push(@b) flattens @b into @a, why should I expect something
different from @a.push(eval(@b.perl))?
> your example is too simple to really cover
> this as you could just push @b or a ref to @b (damn, i need to learn
> more basic p6 syntax! :).
If people want @a.push(\...@b) or @a.push([...@b]), they can just write that -
and if the want to use eval + perl, they can include the brackets or
backslash just as well.
> a more useful example would be serializing data trees. if you dump @b
> with .perl do you want the current dumper output of a anon array or your
> list of values? when serializing a tree, you must get the ref version so
> that is the common and default usage. your version isn't DWIMmy there at
> all.
Maybe we can construct something magic with slice context and captures
instead?
Cheers,
Moritz
Re: Which brackets should @a.perl use?
Uri, On Sun, 04 Jan 2009 22:37:43 -0500, Uri Guttman wrote: > that fails with nested arrays. we don't want them to flatten. > > my $c = eval '(1, (4, 5), 3)'; > > will that work as you envision? No, but it's not what I'm proposing. A reference must Perlify as a reference, just as it does today, so that .perl doesn't destroy information by flattening where you don't want it to. Here's what I propose: my @a = 1, 2, 3; @a.perl.say; # (1, 2, 3) my $ra = @a; $ra.perl.say;# [1, 2, 3] my @b = 1, [2, 3], 4; @b.perl.say; # (1, [2, 3], 4) my $rb = @b; $rb.perl.say;# [1, [2, 3], 4] My objection to the current behaviour is that @a and $ra Perlify to the same string -- '[1, 2, 3]' -- losing the information that @a is an array and $ra is a reference to an array. Therefore, if you serialise and unserialise an array, you always get an array of a single element, containing a reference to the real data. What you get back is not what you put in. Markus
Re: Which brackets should @a.perl use?
> "ML" == Markus Laker writes:
ML> Adding a single backslash before `eval' pushes an anonymous array on to
ML> @b, as you envisage wanting to do:
ML> # Imagine that @a.perl has produced this:
ML> my $p = "('blue', 'light', 'hazard')";
ML> my @b;
ML> @b.push(\eval $p);
but that is manual code. what about a larger tree?
>> a more useful example would be serializing data trees. if you dump @b
>> with .perl do you want the current dumper output of a anon array or your
>> list of values? when serializing a tree, you must get the ref version so
>> that is the common and default usage. your version isn't DWIMmy there at
>> all.
ML> I think Perl 6's automatic reference-taking (though we don't call them
ML> references any more) solves that problem for us.
ML> If you say
ML> my @c = eval '(1, 2, 3)';
ML> then @c has three elements. If you say
ML> my $c = eval '(1, 2, 3)';
ML> then Perl constructs (if I've got the Perl 6 lingo right) an Array object
ML> and stores it in $c. So the round brackets DTRT whether you're storing
ML> into an array like @c or into a scalar like $c.
that fails with nested arrays. we don't want them to flatten.
my $c = eval '(1, (4, 5), 3)';
will that work as you envision? in perl5 with [] it works fine. i know
there are contexts that flatten and others that don't. but a larger tree
with assignments like that are harder to read IMO as lists inside lists
are not nesting but flattening in p5 all the time.
ML> We serialised an array of three elements; we got back an array containing
ML> just one. Round brackets would have solved that. (Actually, we don't
ML> need any brackets at all, because Perl 6's list constructor is a comma,
ML> not a set of brackets. But round brackets would be no-ops, and they
ML> arguably make the output more human-readable.)
try that again with my example above. in p5 the structure would be this:
my $c = [1, [4, 5], 3] ;
how should .perl serialize that so that eval will give back the same
structure? unless () are nesting and not flattening then you can't do it
without a \() which is longer (and uglier IMO than []).
uri
--
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- Perl Code Review , Architecture, Development, Training, Support --
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Re: Which brackets should @a.perl use?
On Sun, 04 Jan 2009 14:19:15 -0500, Uri Guttman wrote:
>> "m" == moritz writes:
> m> But I think that a .perl()ification as ("blue", "light", "hayard",) would
> m> make much more sense, because simple thing like
>
> m> @a.push eval(@b.perl)
>
> m> would then DWIM.
>
> for your def of DWIM. i can see wanting an anon array to be pushed onto
> @a building up a structure.
That would be easily achievable if we made the change I'm suggesting, so
that C<@a.perl> emitted round brackets. This is how you would clone an
array, as Moritz wants to do:
m...@edward:~/perl/6$ cat uri1
#!/home/msl/bin/perl6
# Imagine that @b.perl has produced this:
my $p = "('blue', 'light', 'hazard')";
my @a;
@a.push(eval $p);
.say for @a;
m...@edward:~/perl/6$ ./uri1
blue
light
hazard
m...@edward:~/perl/6$
Adding a single backslash before `eval' pushes an anonymous array on to
@b, as you envisage wanting to do:
m...@edward:~/perl/6$ cat uri2
#!/home/msl/bin/perl6
# Imagine that @a.perl has produced this:
my $p = "('blue', 'light', 'hazard')";
my @b;
@b.push(\eval $p);
.say for @b;
m...@edward:~/perl/6$ ./uri2
blue light hazard
m...@edward:~/perl/6$
> a more useful example would be serializing data trees. if you dump @b
> with .perl do you want the current dumper output of a anon array or your
> list of values? when serializing a tree, you must get the ref version so
> that is the common and default usage. your version isn't DWIMmy there at
> all.
I think Perl 6's automatic reference-taking (though we don't call them
references any more) solves that problem for us.
If you say
my @c = eval '(1, 2, 3)';
then @c has three elements. If you say
my $c = eval '(1, 2, 3)';
then Perl constructs (if I've got the Perl 6 lingo right) an Array object
and stores it in $c. So the round brackets DTRT whether you're storing
into an array like @c or into a scalar like $c.
I'd like to use your example of serialising and unserialising to suggest
that the current behaviour is wrong:
m...@edward:~/perl/6$ cat serialise
#!/home/msl/bin/perl6
my @a = ;
@a.perl.say;
m...@edward:~/perl/6$ cat unserialise
#!/home/msl/bin/perl6
my $p = =$*IN;
my @a = eval $p;
for (@a) { .say }
m...@edward:~/perl/6$ ./serialise | ./unserialise
blue light hazard
m...@edward:~/perl/6$
We serialised an array of three elements; we got back an array containing
just one. Round brackets would have solved that. (Actually, we don't
need any brackets at all, because Perl 6's list constructor is a comma,
not a set of brackets. But round brackets would be no-ops, and they
arguably make the output more human-readable.)
Markus
Re: Which brackets should @a.perl use?
> "m" == moritz writes:
m> S02 says:
m> "To get a Perlish representation of any object, use the .perl method. Like
m> the Data::Dumper module in Perl 5, the .perl method will put quotes around
m> strings, square brackets around list values,"
m> So according to this, Rakudo has it right.
m> But I think that a .perl()ification as ("blue", "light", "hayard",) would
m> make much more sense, because simple thing like
m> @a.push eval(@b.perl)
m> would then DWIM.
for your def of DWIM. i can see wanting an anon array to be pushed onto
@a building up a structure. your example is too simple to really cover
this as you could just push @b or a ref to @b (damn, i need to learn
more basic p6 syntax! :).
a more useful example would be serializing data trees. if you dump @b
with .perl do you want the current dumper output of a anon array or your
list of values? when serializing a tree, you must get the ref version so
that is the common and default usage. your version isn't DWIMmy there at
all.
uri
--
Uri Guttman -- u...@stemsystems.com http://www.sysarch.com --
- Perl Code Review , Architecture, Development, Training, Support --
- Free Perl Training --- http://perlhunter.com/college.html -
- Gourmet Hot Cocoa Mix http://bestfriendscocoa.com -
Re: Which brackets should @a.perl use?
> m...@edward:~/perl/6$ ./ap2
> @c: 3 elements: ["blue", "light", "hazard"]
> @c[0]: blue
> $c: 3 elements: ["blue", "light", "hazard"]
> $c[0]: blue
> m...@edward:~/perl/6$
>
>
> Is Rakudo's behaviour correct here?
S02 says:
"To get a Perlish representation of any object, use the .perl method. Like
the Data::Dumper module in Perl 5, the .perl method will put quotes around
strings, square brackets around list values,"
So according to this, Rakudo has it right.
But I think that a .perl()ification as ("blue", "light", "hayard",) would
make much more sense, because simple thing like
@a.push eval(@b.perl)
would then DWIM.
Chhers,
Moritz
Which brackets should @a.perl use?
This behaviour looks wrong to me:
m...@edward:~/perl/6$ cat ap1
#!/home/msl/bin/perl6
my @a = ;
my $p = @a.perl;
say "\...@a: {...@a.elems} elements: $p";
say '@a[0]: ', @a[0];
my @b = eval $p;
say "\...@b: {...@b.elems} elements: $p";
say '@b[0]: ',@b[0];
say '@b[0][0]: ', @b[0][0];
m...@edward:~/perl/6$ ./ap1
@a: 3 elements: ["blue", "light", "hazard"]
@a[0]: blue
@b: 1 elements: ["blue", "light", "hazard"]
@b[0]: blue light hazard
@b[0][0]: blue
m...@edward:~/perl/6$ perl6 -v
This is Rakudo Perl 6, revision 34744 built on parrot 0.8.2-devel
for i486-linux-gnu-thread-multi.
Copyright 2006-2008, The Perl Foundation.
m...@edward:~/perl/6$
Because C<@a.perl> returns a string surrounded in square brackets, rather
than round brackets, C produces a list containing a single element:
we get one extra, unwanted level of indirection.
If C<@a.perl> were to return a string surrounded in round brackets, this
problem would be solved:
m...@edward:~/perl/6$ cat ap2
#!/home/msl/bin/perl6
my $p = '("blue", "light", "hazard")';
my @c = eval $p;
say "\...@c: {...@c.elems} elements: ", @c.perl;
say '@c[0]: ', @c[0];
my $c = eval $p;
say "\$c: {$c.elems} elements: ", $c.perl;
say '$c[0]: ', $c[0];
m...@edward:~/perl/6$ ./ap2
@c: 3 elements: ["blue", "light", "hazard"]
@c[0]: blue
$c: 3 elements: ["blue", "light", "hazard"]
$c[0]: blue
m...@edward:~/perl/6$
Is Rakudo's behaviour correct here?
Markus
