Re: Perl 6 and Perl 5 parsing

Perl5's grammar is mutable too.

$ perl -E'sub f()  { 0 }  say f + 1;'
1

$ perl -E'sub f($) { 0 }  say f + 1;'
0
[download]

$ perl -E'sub f    { say $_[0]; }  f { foo => "bar" };'
HASH(0x816c158)

$ perl -E'sub f(&) { say $_[0]; }  f { foo => "bar" };'
CODE(0x817bc80)
[download]

$ perl -E'         $x = foo; say $x; sub foo { "sub" }'
foo

$ perl -E'sub foo; $x = foo; say $x; sub foo { "sub" }'
sub
[download]

Update: Added an example to show that prototypes aren't the only source.

Comment on Re: Perl 6 and Perl 5 parsing Select or Download Code

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Re^2: Perl 6 and Perl 5 parsing
by TimToady (Parson) on Jan 14, 2011 at 21:01 UTC

So while Perl 6's grammar is more mutable in theory; it is less frequently mutated by accident.

Another interesting consequence of this decision is that the blocks supplied to map and grep are not a special syntax, so they require a comma. Likewise use of & in a sub signature does not change parsing. On the other hand, bare blocks are just anonymous closures in Perl 6, so they can be passed in any position, and are not restricted to the first argument as they are with Perl 5's & prototype.

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