in reply to Vector manipulation

my $vect1 = $val1->[3.­.5];. When I print test here, I get: 2

Yep. 3..5 is being evaluated in scalar context because dereferencing to get a scalar wants a scalar index: 

 % say "[0..9]->[4,5]"
Useless use of a constant (4) in void context at (eval 1) line 1.
5

[download]

You probably want: 

my $vect1 = [ @{ $val1 }[3.­.5] ];

[download]

- tye        

Replies are listed 'Best First'.
Re^2: Vector manipulation (scalar)
by jcklasseter (Sexton) on Jun 29, 2015 at 12:02 UTC
    Thanks! I got is working on a variant of that, I had to do some dereferencing and re-vectorizing, but it worked! Final snippet: 
            my $vect1 = [ @{ $val1 }[3..5] ];
        my $vect2 = [ @{ $val2 }[3..5] ];
        my $norm = [ @{ $val1 }[0..2] ];
        my $varvec1 = V( @$vect1 );
        my $varvec2 = V( @$vect2 );
        my $varnorm = V( @$norm );
        my $nvect1 = $varvec1 - $varnorm ;
        my $nvect2 = $varvec2 - $varnorm ;

    [download]
[reply]
[d/l]

In Section Seekers of Perl Wisdom