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lisprogtor
lisprogtor

Reputation: 5749

perl6 Passing operators as arguments

I am trying to achieve some functionality similar to Scheme by defining and passing raw operators. In Scheme, you can

(define x +) (x 1 2)

and this operation will give you answer 3. In perl6, I somehow have to place the operator in another function in order to achieve the same:

sub x($a) { $a * 2; }
sub y($m, $n) { $m($n); }
say y(&x, 3); # gives you 6
say y(+, 3) # this is error

Is there an easy way to pass operators and functions?

Thanks.

Upvotes: 2

Views: 145

Answers (1)

Christoph
Christoph

Reputation: 169563

I don't do Scheme, but

(define x +)

should correspond to something like

my constant &x = &infix:<+>;

Then,

say x(1,2)

will output 3 as well.

There's also a shorthand notation &[+] for passing operators around if you find the canonical name used above too cumbersome.

Note that I'm not entirely sure what you were trying to do with your Perl6 code, as that looks rather different from the Scheme snippet.

Upvotes: 6

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