Re: How to find size of array in array?

[jbert]

Good advice in the other responses. Some more words for you:

Your line of code to find the length:

  $asize = @a;

works because an array evaluates to the number of elements it contains in a scalar context. Assigning to a scalar (as you do above) provides a scalar context to the right hand side (RHS). So the array is evaluated in this scalar context, giving the length, which the assignment then puts into your '$asize' variable.

Why did I bother spelling this out in a long-winded way? Because it is useful to understand what it going on. For example, I usually write the above as:

  my $asize = scalar @a;

The 'scalar' keyword supplies a scalar context to its argument and gives the result of the evaluation. It is redundant in the above example (because we already have a scalar context from the assignment to the scalar).

So why do I write it like that? Because:

• It is an explicit statement of what I am doing. I consider that more readable.
• I can use the same syntax every place I need an array size. Look at the following:

  #!/usr/local/bin/perl -w
  use strict;
  
  my @array = qw( 1 2 3 4 5 );
  my $asize = scalar @array;
  print "Array size 1 is ", $asize, "\n";
  print "Array size 2 is ", @array, "\n";
  print "Array size 3 is ", scalar @array, "\n";
  

  I *need* to use the scalar keyword here, because otherwise the 'print' command gives a list context to the array, which lists all the elements instead of giving the size.

  Since I sometimes get cut-and-paste happy, it is nice to have code which is explicit about what context it wants.

And I'll conclude this overly-pedantic missive to state that you should look into using 'use strict' and '#!... -w', to force you into good habits. (Heh...habits! Monks! How funny is that!? (Hrrrm...the first couple of million times were probably worth chuckling at, but thats about it))