On 07/03/2015 10:02 PM, yary wrote:
> On Fri, Jul 3, 2015 at 3:03 PM, Timo Paulssen wrote:
>> but this does not
>>
>>> sub takes_int_array(Int @bar) { say @bar }
>>> takes_int_array([1, 2, 3])
>>
>> because the type match is against the defined type. We do not
>> automatically infer that [1, 2,
On 07/03/2015 10:02 PM, yary wrote:
> On Fri, Jul 3, 2015 at 3:03 PM, Timo Paulssen wrote:
>> but this does not
>>
>>> sub takes_int_array(Int @bar) { say @bar }
>>> takes_int_array([1, 2, 3])
>> because the type match is against the defined type. We do not
>> automatically infer that [1, 2, 3] co
On Fri, Jul 3, 2015 at 3:03 PM, Timo Paulssen wrote:
> but this does not
>
>> sub takes_int_array(Int @bar) { say @bar }
>> takes_int_array([1, 2, 3])
>
> because the type match is against the defined type. We do not
> automatically infer that [1, 2, 3] could be a Positional[Int].
How would one c
On Fri, Jul 3, 2015 at 2:03 PM, Timo Paulssen wrote:
> On 07/03/2015 07:20 PM, Tom Browder wrote:
>> On Fri, Jul 3, 2015 at 10:26 AM, Tom Browder wrote:
...
>> What is the proper type to use for the %hash for a more complete
>> signature in function foo1? I've tried various types such
...
> When
On 07/03/2015 07:20 PM, Tom Browder wrote:
> On Fri, Jul 3, 2015 at 10:26 AM, Tom Browder wrote:
>> While experimenting I've found the first two methods of passing a hash
>> to a subroutine work:
>>
>> # method 1
>> my %hash1;
>> foo1(%hash1);
>> say %hash1.perl;
>> sub foo1(%hash) {
>> %hash{1}
On Fri, Jul 3, 2015 at 10:26 AM, Tom Browder wrote:
> While experimenting I've found the first two methods of passing a hash
> to a subroutine work:
>
> # method 1
> my %hash1;
> foo1(%hash1);
> say %hash1.perl;
> sub foo1(%hash) {
> %hash{1} = 0;
> }
Another question on method 1 above, please:
On Jul 3, 2015 11:14 AM, "Brandon Allbery" wrote:
>
> On Fri, Jul 3, 2015 at 11:26 AM, Tom Browder
wrote:
>>
>> # method 1
...
>> foo1(%hash1);
> This is what I would naïvely expect to work in any language except Perl 5.
That comment, along with Liz's, has convinced me to use method 1 (less
typi
On Fri, Jul 3, 2015 at 11:26 AM, Tom Browder wrote:
> # method 1
> my %hash1;
> foo1(%hash1);
> say %hash1.perl;
> sub foo1(%hash) {
> %hash{1} = 0;
> }
>
This is what I would naïvely expect to work in any language except Perl 5.
> # method 2
> my %hash2;
> my $href2 = %hash2;
> foo2($href2)
> On 03 Jul 2015, at 17:26, Tom Browder wrote:
>
> While experimenting I've found the first two methods of passing a hash
> to a subroutine work:
>
> # method 1
> my %hash1;
> foo1(%hash1);
> say %hash1.perl;
> sub foo1(%hash) {
> %hash{1} = 0;
> }
>
> # method 2
> my %hash2;
> my $href2 = %ha
While experimenting I've found the first two methods of passing a hash
to a subroutine work:
# method 1
my %hash1;
foo1(%hash1);
say %hash1.perl;
sub foo1(%hash) {
%hash{1} = 0;
}
# method 2
my %hash2;
my $href2 = %hash2;
foo2($href2);
say %hash2.perl;
sub foo2($href) {
$href{1} = 0;
}
# thi
10 matches
Mail list logo
