Got it, thanks.
Xin
> On Jun 9, 2018, at 4:07 PM, Brandon Allbery wrote:
>
> And in the others, you've provided an explicit invocant with " some kind>.sort", so again it knows it's a method call and has an invocant
> already.
>
> A sub can be forced to be a method call instead by using ":"
And in the others, you've provided an explicit invocant with ".sort", so again it knows it's a method call and has an invocant
already.
A sub can be forced to be a method call instead by using ":" and providing
the invocant *before* the colon: say sort(<3 5 2 1>: {$^a <=> $^b})
On Sat, Jun 9,
Thanks. But I am actually confused by the use of colon in
Sort: { ... }
What does it mean in the above statement? I have done several experiments like:
p6 'say sort({$^a <=> $^b}, < 3 5 2 1>)'# (1 2 3 5)
p6 'say <3 5 2 1>.sort({$^a <=> $^b})' # it works.
p6 'say <3 5 2
The ".=" operator means call the method on the right, with the thing on the
left as invocant, and assign the result back to the thing on the left. So
@x .= sort: ...
is the same as
@x = @x.sort(...)
So you're being confused by the syntactic "magic" of ".=".
On Sat, Jun 9, 2018 at 2:58
I got the point for //.
Another question is about calling the method sort with a code block. I can
understand
@x .= sort({ ... });
But I don't quite understand why this form also works.
@x .= sort: { ... };
I look into the documentation for infix ":", https://docs.perl6.org/routine/:
More precisely, at that point you have a bunch of numbers, but possibly not
as many as expected if some of the components weren't numeric (or all of
them, as when there are files present that aren't the expected logs). Which
means some or all of those variables will be undefined instead of
The magic trick is that the last line in the code block is the return value.
when you pass a function that takes a single argument (in this case just
the $_) to sort, it will call your function for every element in the
input array and use the result of that function to sort the elements by.
The
