Yes, there is an absolutely tiny mention of it on the page: Metaoperators
Metaoperators can be parameterized with other operators or subroutines in the same way as functions can take functions as parameters. To use a subroutine as a parameter, prefix its name with a &. Perl 6 will generate the actual combined operator in the background, allowing the mechanism to be applied to user defined operators. *To disambiguate chained metaoperators, enclose the inner operator in square brackets.* There are quite a few metaoperators with different semantics as explained, next. If that's even the part you're asking about, that is. And yes, I would consider that LTA, making it easier to find, for example by adding an example or maybe even a sub-heading, would be really good. Who wants to open a ticket? :) If you want an example of a disambiguated metaop, try something like Z+= which could either be [Z+]= or Z[+=]. On 06/10/2018 00:36, Trey Harris wrote: > > On Fri, Oct 5, 2018 at 8:33 AM Timo Paulssen t...@wakelift.de > <http://mailto:t...@wakelift.de> wrote: > > It's important to point out that inside metaoperators ("composed > operators", "combined operators", ...) the [ ] are just for bracketing > things together — sometimes it's needed to disambiguate, but you > can put > it in anyway to make things clearer. > > This has nothing to do with [+] 1, 2, 3, which is a way to spell "add > these three numbers together". > > Timo—is this fact formalized in the docs? I can’t find it. > https://docs.perl6.org/routine/= doesn’t mention it, and while > https://docs.perl6.org/language/operators#Assignment_operators gives > an example of using it (with |+=|, |-=|, |min=| and |~=|), I can’t > find the metaoperator version’s definition. > > > On 05/10/2018 14:26, Brad Gilbert wrote: > > Note that = is actually a meta operator that can take an infix > > operator as an argument > > > > So > > > > $a += 1 > > > > is really short for > > > > $a [+]= 1 > > > > On Fri, Oct 5, 2018 at 1:02 AM Todd Chester > <toddandma...@zoho.com <mailto:toddandma...@zoho.com>> wrote: > >> > >> > >> On 10/4/18 12:13 PM, Brandon Allbery wrote: > >>> It's fine syntactically, but I think has no effect because > it'd be '$v = > >>> $v' after the '+|='. Conceivably some future version of rakudo > could > >>> warn about it having no effect. > >> That explains it. Thank you! >
