On 10/12/18 2:35 PM, Curt Tilmes wrote:
On Fri, Oct 12, 2018 at 5:08 PM ToddAndMargo via perl6-users
<perl6-us...@perl.org <mailto:perl6-us...@perl.org>> wrote:
>> On 10/12/18 12:52 PM, Curt Tilmes wrote:
>> > You could make a subset for the List your're trying to
return:
>> >
>> > subset liststrint of List where .[0] ~~ Str && .[1] ~~ Int;
>> > sub RtnOrd( Str $Char --> liststrint) ...
>>
>> I am confused.
>>
>> I want to get the --> syntax correct for `return $Char,
ord($Char)`
On 10/12/18 1:49 PM, Brad Gilbert wrote:
> That would be `List`
>
> sub RtnOrd( Str $Char --> List ){ $Char, ord($Char) }
> say RtnOrd "A"
> # (A 65)
$ p6 'sub RtnOrd( Str $Char --> List ){return $Char, ord($Char)}; say
RtnOrd "A";'
(A 65)
But "List" does not tell my what is in the list.
You can create a brand new type, a subset of Lists where the first element
(we refer to with [0]) is of type Str (~~ Str) and the second element of
the List
(we refer to with [1]) is of type Int (~~ Int).
Define it like this:
subset list-str-int of List where .[0] ~~ Str && .[1] ~~ Int;
then you can say that your routine returns a list that looks like that:
sub RtnOrd( Str $Char --> list-str-int)
Is there any way to say I am return two things: a string and an integer?
