Re^2: @array elements multiplication with another array elements.

[grizzley]

in reply to Re: @array elements multiplication with another array elements.
in thread @array elements multiplication with another array elements.

Wow, how can pairwise tell where the end of 1st array is? There was always the problem feature that regardless whether you write

my @array1 = (3,4,6,5,8);
my @array2 = (2,2,4,5,1);

or

my @array1 = (3,4,6,5,8,2,2,4);
my @array2 = (5,1);

using it with coma always results in flattened list:

@array1,@array2

is equivalent to

(3,4,6,5,8,2,2,4,5,1)

and yet pairwise deals with it:

c:\>perl -le "use List::MoreUtils qw/pairwise/; @a=(1..5);@b=(11..15);
+ print for pairwise {$a+$b} @a, @b"
12
14
16
18
20

c:\>perl -le "use List::MoreUtils qw/pairwise/; @a=(1..5, 11,12);@b=(1
+3..15); print for pairwise {$a+$b} @a, @b"
14
16
18
4
5
11
12

[tobyink]

It uses a prototype of &\@\@. That means the two arrays get implicitly turned into arrayrefs by the Perl parser. The body of the sub just sees a coderef and two arrayrefs - not a flattened list. See perlsub.

[grizzley]

That's true! Wow, thanks, first example of prototypes being useful

c:\>perl -le "sub f(\@\@) { print 'array ', ++$cnt ,': ', join ' ', @$
+_ for @_ } @a=(1..5, 11,12);@b=(13..15); f @a, @b"
array 1: 1 2 3 4 5 11 12
array 2: 13 14 15