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

In reply to Re^2: @array elements multiplication with another array elements. by grizzley
in thread @array elements multiplication with another array elements. by New Perlmonk

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