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;
}
#
On 03 Jul 2015, at 17:26, Tom Browder tom.brow...@gmail.com 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
On Jul 3, 2015 11:14 AM, Brandon Allbery allber...@gmail.com wrote:
On Fri, Jul 3, 2015 at 11:26 AM, Tom Browder tom.brow...@gmail.com
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
On Fri, Jul 3, 2015 at 10:26 AM, Tom Browder tom.brow...@gmail.com 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
On Fri, Jul 3, 2015 at 2:03 PM, Timo Paulssen t...@wakelift.de wrote:
On 07/03/2015 07:20 PM, Tom Browder wrote:
On Fri, Jul 3, 2015 at 10:26 AM, Tom Browder tom.brow...@gmail.com wrote:
...
What is the proper type to use for the %hash for a more complete
signature in function foo1? I've
On 07/03/2015 10:02 PM, yary wrote:
On Fri, Jul 3, 2015 at 3:03 PM, Timo Paulssen t...@wakelift.de 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]
