Another article of the series “Yet Another Perl 6 Operator”

Among the new Perl 6 operators, there is the handy operator '//', known as defined-or or the default operator. This novelty was anticipated by the introduction of this syntactic bit in Perl 5 (see the upcoming 5.10 release) — so you won’t need to wait for Perl 6 to start using it.

# dor.pl
use 5.010;
print "arg: '", shift // "?", "'\n";

$ perl dor.pl one
arg: 'one'
$ perl dor.pl ""
arg: ''
$ perl dor.pl
arg: '?'

The purpose of this operator is very close to the well-known high-precedence OR, the operator '||', with some subtle and useful differences.

Expressions with '||' are evaluated like this:

exp1 || exp2  =  exp1,   if the result of exp1 is true
exp1 || exp2  =  exp2,   otherwise

It was very common to use this operator when one wanted an expression to have a certain default value if something was not explicitly specified.

For instance, the following Perl 5 code can build a rule of length $n with a given char or '-'.

sub rule {
    my $n = shift;
    my $char = shift || '-';
    return $char x $n;
}

Then 

rule(3)         gives   '---'
rule(4, '+')    gives   '++++'

But 

rule(5, 0)      gives   '-----' (and not '00000')

because 0 is false (as the other Perl values of falsehood: ‘’ – the empty string, 0 – integer zero, 0.0 – floating point zero, undef).

The solution was to replace the simple expression shift || '-' by another one, more complex and precise: 

my $char = defined $_[0] ? shift : '-';

But that is the semantics of the '//' operator:

exp1 // exp2  =  exp1,   if the result of exp1 is defined
exp1 // exp2  =  exp2,   otherwise

So since 5.10, you may use the exact same code above replacing '||' with '//' and keep your code compact and elegant.

To be really fair, the Perl 6 rephrasing of this code won’t need the '//' operator. Instead, that sub would be even prettier due to the improved sub signatures that become possible. 

sub rule ( $n, $char = '-' ) {
  return $char x $n;
}

Next article is due tomorrow.


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