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Talk:Call a function: Difference between revisions
→Operators: yes, you can redefine any p6 operator in a lexical scope
m (Hey, I found some notes entitled "Metaoperators".) |
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::Ok, cheers Tim. If that is the case, then I agree it was right to document it here. Like I say I don't know the language. Out of interest, if an operator is a function, can it be redefined? Is there an example of say, a redefinition of an addition function that is used for the addition operator? I plan to learn Perl 6 in future, when the compiler rolls down into Debian stable. [[User:Markhobley|Markhobley]] 06:34, 18 July 2011 (UTC)
:::Hey, I found some notes entitled "Metaoperators", that explains this operator/function stuff. Cheers all. [[User:Markhobley|Markhobley]] 06:42, 18 July 2011 (UTC)
::Yes, that's one of the nice things about treating operators as functions: they end up lexically scoped just like functions, and obey the same shadowing and/or multiple-dispatch rules. So if you define your own <tt>infix:<+></tt> it can either override or cooperate with any outer definitions of that operator. And even the metaoperators are just calls to higher-order functions underneath. These translations happen very early in the compilation; in a sense, all operators in Perl 6 are just convenient macros that rewrite the AST to a purer FP and/or OO form, and then either the early-binding lexical rules or the late-binding inheritance rules control which function gets called. We're pretty happy with how simple this foundation turned out, given everything we're trying to build on top of it.
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