We don't bite newbies here... much PerlMonks

### Re^4: RFC: Is there more to alias?

by xmath (Hermit)
 on Aug 24, 2004 at 23:12 UTC ( #385541=note: print w/replies, xml ) Need Help??

in reply to Re^3: RFC: Is there more to alias?
in thread RFC: Is there more to alias?

Doesn't something have to be aliased to something else? What is being aliased to what in these examples?

\$x = alias [\$y, \$z];

\$x->[0] is an alias to \$y and \$x->[1] is an alias to \$z.

alias push @x, \$y;
\$x[-1] is an alias to \$y.

Replies are listed 'Best First'.
Re^5: RFC: Is there more to alias?
by fergal (Chaplain) on Aug 24, 2004 at 23:46 UTC
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!

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

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:-(

Create A New User
Node Status?
node history
Node Type: note [id://385541]
help
Chatterbox?
and all is quiet...

How do I use this? | Other CB clients
Other Users?
Others scrutinizing the Monastery: (5)
As of 2017-07-28 16:54 GMT
Sections?
Information?
Find Nodes?
Leftovers?
Voting Booth?
I came, I saw, I ...

Results (431 votes). Check out past polls.