Carl Mäsak wrote:
Andy (>):
map?
perl6 -e 'my $x = :a<5>; say $x.map( { .value / 10} ).fmt("%s")'
Yes, sure. That'll print a tenth of the value of $x. The '.fmt("%s")'
is a no-op in this case.
// Carl
Not entirely a no-op. Thus
$perl6
> my $x=:a<5>; say $x.map({.value/10}).fmt("
On Thu, Jan 15, 2009 at 12:46:09PM +0300, Richard Hainsworth wrote:
> Larry Wall wrote:
>> On Wed, Jan 14, 2009 at 09:55:38AM +0300, Richard Hainsworth wrote:
>>
>>> However, I came across one thing in solution #3 that I posted
>>> yesterday. $pair.fmt("%s %s") is nice, but it doesnt allow for
Larry Wall wrote:
On Wed, Jan 14, 2009 at 09:55:38AM +0300, Richard Hainsworth wrote:
However, I came across one thing in solution #3 that I posted yesterday.
$pair.fmt("%s %s") is nice, but it doesnt allow for any action on either
value or key before printing (I wanted to print the value
On Wed, Jan 14, 2009 at 09:55:38AM +0300, Richard Hainsworth wrote:
> However, I came across one thing in solution #3 that I posted yesterday.
> $pair.fmt("%s %s") is nice, but it doesnt allow for any action on either
> value or key before printing (I wanted to print the value as a
> percenta
Richard (>), Carl (>>), Andy (>>>):
>>> P6 treats the key/value as an anonymous 'pair' object so @ranking is an
>>> list of pairs. That's why:
>>> say @ranking.pop.fmt("$m Medal: %s, %s")
>>>
>>> or, less succinctly:
>>> say (pop @ranking).fmt("$m Medal: %s, %s");
>>>
>>> works - the pair object,
Carl Mäsak wrote:
Andy (>):
P6 treats the key/value as an anonymous 'pair' object so @ranking is an
list of pairs. That's why:
say @ranking.pop.fmt("$m Medal: %s, %s")
or, less succinctly:
say (pop @ranking).fmt("$m Medal: %s, %s");
works - the pair object, popped off into the 'printf' l
Of course whenever I post to this list I make some sort of mistake, like
including a test version that does not parse in perl6. Hopefully, I have
eliminated all the mistakes.
The following is a revised version due to Carl Masak.
The problem and model perl5 solution by Jan Dubois (a member of th
