Re^9: Definition of numerically equal and rationale for 'you' == 'me'

[moritz]

in reply to Re^8: Definition of numerically equal and rationale for 'you' == 'me'
in thread Definition of numerically equal and rationale for 'you' == 'me'

Well, everything is possible with the right amount of disambiguation rules. Perl 6 doesn't have barewords, and type names must be predeclared. If a subroutine of name x and another symbol with that name (type name, constant, enum element) are in scope, the parser disambiguates in favor of the latter, unless there are parens after the token.

In fact, constants can give you a taste of sigil-less Perl 6 already:

use v6;
constant x = 4; 
say x x x # produces 4444

This works today in both Rakudo and Niecza.

[JavaFan]

That works in Perl 5 as well:

perl -Mconstant=x,1 -MO=Deparse -e 'print x x x'
print '1' x '1';
-e syntax OK

But:

perl  -MO=Deparse -e 'sub x {1} print x x x'
sub x {
    1;
}
print x(x(x()));
-e syntax OK

[moritz]

Exactly. That's what I meant with my last paragraph in Re^7: Definition of numerically equal and rationale for 'you' == 'me'. Without sigils, you have to be much more careful, about whether a term or an infix is expected, and errors are more likely to go unnoticed. So possible, but in general a bad idea.