Re^5: RFC: Is there more to alias?

[fergal]

Does the first 1 autovivify the arrayref in $x (as $x->[0] = 1 would) or must $x already have an array ref in it? What happens if I do shift @$x? Is $x->[0] now an alias to $z and the alias to $y disappears? Or (unlikely) does it effectively do @{$x} = @{$x}[1..($#$x-1)] which amounts to

$y = $z;
$z = $x->[2];
$x->[2] = $x->[3];
....
$x->[$#$x - 1] = $x->[$#$x];
pop @$x;

Fun and games!

The second example one seems very odd to me. To me, it reads as alias(push(@x, $y)) but I guess perl doesn't parse it as that. Why do this rather than just alias $x[-1] = $y or should that be alias $y = $x[-1]?

So you can use , or = to separate the aliased things? And you can put alias on either side of the =?

Anyway, time for bed. I dreamt about object pascal last night, I hope I don't dream about aliases to night!

[xmath]

You still misundestand how Data::Alias works, and you seem to see too much magic where there is none. Data::Alias does no parsing, perl does the parsing, so all syntax is just perl's syntax. Like I said in the description, "alias" itself is a nop. To perl, it's thin air. In fact, if you look at the operations executed (-MO=Concise,-exec), for example for alias push @x, $y:

3  <0> pushmark s
4  <#> gv[*x] s
5  <1> rv2av[t3] lKRM/1
6  <#> gvsv[*y] sM
7  <@> push[t5] vK/2
-  <1> ex-list vK

See the call to alias? Nope, it isn't even there, apart from a miniscule stub "ex-list" (a nop). Alias just changes the semantics of whatever is inside it, and in a very simple way: whereever perl normally copies data, aliasing occurs instead.

So, let's analyze the cases. Take $x = alias [$y, $z] for example. What does it do? The [$y, $z] is just the array constructor which normally creates a new array, fills it with copies of $y and $z, and returns a reference to the array. Within the scope of "alias", it therefore creates a new array, fills it with aliases to $y and $z, and returns a reference to the array, which it then normally assigned to $x.

alias push @x, $y is indeed as you noted parsed as alias(push(@x, $y)). This push would normally add a copy of $y onto the end of @x, so within "alias" it adds an alias to $y onto the end of @x. This obviously differs from alias $x[-1] = $y which would overwrite the last element of @x with an alias to $y.

You finally mentioned using a comma... ($x, $y) involves no copying, therefore alias($x, $y) does nothing remarkable. It behaves 100% identical to plain ($x, $y).

All cleared up now? :-)

Time for me to get some sleep too.. zZ

[fergal]

I didn't think you were parsing anything but I misunderstood what alias does and so assumed that my parsing of alias push @x, $y was wrong. I'm used to thinking of "alias this to that" rather than "turn on aliasing for this bit of code".

I know you changed the lexer to make this unnecessary but I think syntax like

alias { push(@x, y) }

would give a clearer indication of what is going on. It would also cause problems with scoping when you want to use my:-(

[xmath]

You can use either syntax, so if you think alias { push(@x, $y) } is clearer, use that. (I'm personally a fan of sparse use of parens and braces, but maybe that's just my strange taste). Note however indeed that if you try to do my-declarations, you can run into trouble... alias { my $x = $y } would do nothing useful, since the scope of $x is limited to the block.

On a side note, the lexer hack is actually to make alias { .. } work. alias LIST is the normal mode of operation.