Re^7: Definition of numerically equal and rationale for 'you' == 'me'

[moritz]

I suppose the vast number of operators that Perl supports, compared to other languages, is made possible by sigils; that is, languages without sigils couldn't so easily use x or xx as an operator.

Technically that's not quite correct. The really interesting property that makes that possible is predictive parsing. That is, the Perl parsers always knows whether to expect an infix operator or something termish (a prefix operator or a term, like for example a number, variable, subroutine call etc.).

Even in an imaginary Perl without sigils, the following code could be parsed unambiguously:

# imaginary sigilless Perl
my x = 42;
say x x x;

The parser knows that say is a listop, and after that it expects a term. Thus the first x is term (here a variable), but since two terms in a row are forbidden, the next thing must be an infix operator (or maybe a semicolon to terminate the statement). Thus the second x is parsed as the repetition operator. And so on.

Predictive parsing is also what makes it possible to use the slash / both as an infix operator, and as a regex delimiter.

But even if it worked in theory, a sigil-less language with such a wealth of infix operators would probably be rather hard to read, and it would be much harder (if not impossible) to give good error messages in the case of syntax errors.

[JavaFan]

Even in an imaginary Perl without sigils, the following code could be parsed unambiguously:

No, it cannot. Stepping away the additional problem with say (the first x can be seen as a filehandle), you're forgetting that sigils make barewords possible. You're x x x can be parsed in several ways:

$x x $x
x(x($x))
x::->x("x")

or some combination there off. That is, if we'd dropped sigils, a bare x can be a (scalar) variable, an operator, a class name, a subroutine, or the string "x".

[moritz]

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