Re: Perl Cannot Be Parsed: A Formal Proof

I'm confused by this proof and I don't buy it, especially since the definitions, including that of "parsing," are too imprecise to warrant some reduction proof. You also don't need reduction to show the existence of an example, which is what you're essentially trying to do. You seem to be making the claim that when you have code

  BEGIN {
    x();
    sub foo { }
  }
[download]

then foo() will be declared only when x() terminates.

But that's like writing this Java code

  for(;;) { }
  int x;
[download]

and asking whether "x" will ever be declared, which says absolutely nothing about how this program is parsed (and it's parsed fine).

Cheers.

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Re^2: Perl Cannot Be Parsed: A Formal Proof by Jeffrey Kegler (Hermit) on Aug 15, 2009 at 01:53 UTC
In a series of three articles in The Perl Review, now available online, I laid this proof out much more carefully, in three different versions. The most bullet-proof version derives the result directly from Rice's Theorem. You are right that the proof would not go through if the only way of establishing a prototype for a Perl function was via a function definition. As I said in Perl Review (4.3, p. 28): I must show that I can use Turing-complete Perl code to determine the prototype of the `dunno` subroutine at compile time. A function definition won’t work. Summarizing from that article, there are at least two ways that the prototype of a subroutine can be established using Turing-complete Perl code. One is with symbol table manipulation, for example: `BEGIN { *dunno = sub () { 3 } }` A second way is to put the function definition into a string which is `eval`'d in a BEGIN block. I believe clever monks will be able to think of others.	[reply] [d/l]
Re^2: Perl Cannot Be Parsed: A Formal Proof by chromatic (Archbishop) on Aug 14, 2009 at 02:31 UTC
You seem to be making the claim that when you have code `BEGIN { x(); sub foo { } }` [download] then `foo()` will be declared only when x() terminates. That particular example has a flaw, but it's easy to demonstrate that the argument holds: `use Modern::Perl; sub x { say "In x!" } BEGIN { x() } sub foo!bar { say 'In foo!bar!' } BEGIN { foo!bar() }` [download] For extra fun, run it through B::Deparse.	[reply] [d/l] [select]