See also perllol, perlreftut.

Thank you for your help. It didn't even occur to me to use push.

A "two dimensional array" is really a one dimensional array whose entries are ref:ARRAY, that is references to arrays.

So you want something along the lines of:

  push @sheets, [split /\s+/, $line] ;
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BTW, I wasn't sure whether you meant split /(\s)+/, which I believe will capture the whitespace separators as items in the resulting list ?

... and if /(\s)+/ is really meant, it is cleaner written as /\s*(\s)/, in which case it's obvious that only the last matched whitespace will actually be in $1.

I as having a seperate issue with the whitespace. Your suggestion to remove the parens solved in perfectly. Also, the syntax for the push is very concise. Thank you for your help. Joe

$sheets[1] = \@column;
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Perhaps?

No one can tell you for sure. They can only guess, just as I have.

Can you some show us some data, and what you expect to happen with it?

$sheets[1] = \@column;

Perhaps?

As a general proposition, I would worry about this construct -- in particular I would worry whether @column could be changed later, generating side effects in the @sheet array.

However, the following works fine:

  my @sheet = () ;
  while ($line = <SHEET>) {
    my @columns = split /\s+/, $line ;
    push @sheet, \@columns ;
  } ;
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Nevertheless, alarm bells might be going off in the back of one's mind: this appears to be assigning a pointer to an "automatic" variable (eek!).

It's OK though, you just have to remember that @columns is not really the name of the array, it's the name of a pointer to an array data structure on the heap; and, \@columns isn't returning the address of @columns, it's making a copy of the pointer. When @columns drops out of scope, it's copy of the pointer disappears, but the array data structure stays in the heap until all pointers to it disappear. Obvious, really :-)

The paranoid might worry that @columns might have a life-time of the while loop -- so we'd have to worry about what the assignment does ! Though Perl is, at heart, an interpretted language, so my is an executable item, which might reinitialise @columns each time around the loop, before the assignment ?

It would be easy to make this mistake:

  my (@sheet, @columns) ;
  while (my $line = <SHEET>) {
    @columns = split /\s+/, $line ;
    push @sheet, \@columns ;
  } ;
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which does leave @sheet with all entries pointing to the same thing: the last line of the input ! (From which we deduce that the assignment to @columns overwrites the array data structure that @columns implicitly points to. Which in some ways is a bit of a surprise ! We also deduce that in the previous case, my @columns within the while loop was reset before the assignment, or was discarded at the end of the loop each time around.)

Anyway, the point here is: when constructing data structures you can use \@foo or \%bar, but in doing so you (a) have to be careful, and (b) have to know what it is you have to be careful about !

  push @sheet, [ @columns ] ;
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is unambiguous and secure -- but I guess requires a copy of the array data to be made. One hopes that:

  push @sheet, [ split /\s+/, $line ] ;
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is bright enough to spot that the list produced by split is already anonymous, so that the [...] needs only to return its address !

As a general proposition, I would worry about this construct -- in particular I would worry whether @column could be changed later, generating side effects in the @sheet array.

You seem to assume that isn't what the OP wanted. Sometimes that's exactly why \@columns must be used over [ @columns ].

The OP was sufficiently vague that neither of us could have guessed at what they wanted.

The syntax that creates the copy, @columns , was sufficient for what I need to do in this case (basically iterate through a text file and read each space delimited value into a multi-dimensional array). However, it's good to know that these array pointers can be controled so precicely. Never knew that you could assign a 2d array by value or by reference, so to speak. Thank you for making this distinction. I'll definately reference this post when I need to create it as a pointer. I appreciate the help.

push @sheet,[@array];
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$sheet[$n]=[@array];
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push @sheet,\@array;
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$sheet[$n]=\@array;
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I like this break-down of the different ways to accomplish this. Thank you.