Avoiding Agnostic Jagged Array Flattening in Powershell

Question

Hello,

I'm running into an interesting problem in Powershell, and haven't been able to find a solution to it. When I google (and find things like this post), nothing quite as involved as what I'm trying to do comes up, so I thought I'd post the question here.

The problem has to do with multidimensional arrays with an outer array length of one. It appears Powershell is very adamant about flattening arrays like @( @('A') ) becomes @( 'A' ). Here is the first snippet (prompt is >, btw):

> $a =  @( @( 'Test' ) )
> $a.gettype().isarray
True
> $a[0].gettype().isarray
False

So, I'd like to have $a[0].gettype().isarray be true, so that I can index the value as $a[0][0] (the real world scenario is processing dynamic arrays inside of a loop, and I'd like to get the values as $a[$i][$j], but if the inner item is not recognized as an array but as a string (in my case), you start indexing into the characters of the string, as in $a[0][0] -eq 'T').

I have a couple of long code examples, so I have posted them at the end. And, for reference, this is on Windows 7 Ultimate with PSv2 and PSCX installed.

Consider code example 1: I build a simple array manually using the += operator. Intermediate array $w is flattened, and consequently is not added to the final array correctly. I have found solutions online for similar problems, which basically involve putting a comma before the inner array to force the outer array to not flatten, which does work, but again, I'm looking for a solution that can build arrays inside a loop (a jagged array of arrays, processing a CSS file), so if I add the leading comma to the single element array (implemented as intermediate array $y), I'd like to do the same for other arrays (like $z), but that adversely affects how $z is added to the final array.

Now consider code example 2: This is closer to the actual problem I am having. When a multidimensional array with one element is returned from a function, it is flattened. It is correct before it leaves the function. And again, these are examples, I'm really trying to process a file without having to know if the function is going to come back with @( @( 'color', 'black') ) or with @( @( 'color', 'black'), @( 'background-color', 'white') )

Has anybody encountered this, and has anybody resolved this? I know I can instantiate framework objects, and I'm assuming everything will be fine if I create an object[], or a list<>, or something else similar, but I've been dealing with this for a little bit and something sure seems like there has to be a right way to do this (without having to instantiate true framework objects).

Code Example 1

function Display($x, [int]$indent, [string]$title)
{
  if($title -ne '') { write-host "$title`: " -foregroundcolor cyan -nonewline }
  if(!$x.GetType().IsArray)
  { write-host "'$x'" -foregroundcolor cyan }
  else
  {
    write-host ''
    $s = new-object string(' ', $indent)
    for($i = 0; $i -lt $x.length; $i++)
    {
      write-host "$s[$i]: " -nonewline -foregroundcolor cyan
      Display $x[$i] $($indent+1)
    }
  }
  if($title -ne '') { write-host '' }
}


### Start Program
$final = @( @( 'a', 'b' ), @('c'))
Display $final 0 'Initial Value'

### How do we do this part ??? ###########
                                        ##
$w = @( @('d', 'e') )                   ##
$x = @( @('f', 'g'), @('h') )           ##
# But now $w is flat, $w.length = 2     ##
                                        ##
                                        ##
# Even if we put a leading comma (,)    ##
# in front of the array, $y will work   ##
# but $w will not. This can be a        ##
# problem inside a loop where you don't ##
# know the length of the array, and you ##
# need to put a comma in front of       ##
# single- and multidimensional arrays.  ##
$y = @( ,@('D', 'E') )                  ##
$z = @( ,@('F', 'G'), @('H') )          ##
                                        ##
                                        ##
##########################################

$final += $w
$final += $x
$final += $y
$final += $z
Display $final 0 'Final Value'


### Desired final value: @( @('a', 'b'), @('c'), @('d', 'e'), @('f', 'g'), @('h'), @('D', 'E'), @('F', 'G'), @('H') )
### As in the below:
# 
# Initial Value:
# [0]:
#  [0]: 'a'
#  [1]: 'b'
# [1]:
#  [0]: 'c'
# 
# Final Value:
# [0]:
#  [0]: 'a'
#  [1]: 'b'
# [1]:
#  [0]: 'c'
# [2]:
#  [0]: 'd'
#  [1]: 'e'
# [3]:
#  [0]: 'f'
#  [1]: 'g'
# [4]:
#  [0]: 'h'
# [5]:
#  [0]: 'D'
#  [1]: 'E'
# [6]:
#  [0]: 'F'
#  [1]: 'G'
# [7]:
#  [0]: 'H'

Code Example 2

function Display($x, [int]$indent, [string]$title)
{
  if($title -ne '') { write-host "$title`: " -foregroundcolor cyan -nonewline }
  if(!$x.GetType().IsArray)
  { write-host "'$x'" -foregroundcolor cyan }
  else
  {
    write-host ''
    $s = new-object string(' ', $indent)
    for($i = 0; $i -lt $x.length; $i++)
    {
      write-host "$s[$i]: " -nonewline -foregroundcolor cyan
      Display $x[$i] $($indent+1)
    }
  }
  if($title -ne '') { write-host '' }
}



function funA()
{
  $ret = @()
  $temp = @(0)
  $temp[0] = @('p', 'q')
  $ret += $temp
  Display $ret 0 'Inside Function A'
  return $ret # What about return ,$ret ? What about if $ret = @( @('p', 'q'), @('r', 's') ) -- would return ,$ret still work?
}

function funB()
{
  $ret = @( ,@('r', 's') )
  Display $ret 0 'Inside Function B'
  return $ret
}


### Start Program

$z = funA
Display $z 0 'Return from Function A'
$z = funB
Display $z 0 'Return from Function B'


### Desired final value: @( @('p', 'q') )  and same for r,s
### As in the below:
# 
# Inside Function A:
# [0]:
#  [0]: 'p'
#  [1]: 'q'
# 
# Return from Function A:
# [0]:
#  [0]: 'p'
#  [1]: 'q'

Thanks, Matt

Answer 1

There is another question that starts with the same problem: http://stackoverflow.com/questions/803521/powershell-pitfalls . It looks like this is done by design.

I think if you return ,$ret instead of $ret, it should work.

Two more notes:

• you can test, if the item is array by $item -is [array] (just because it looks more like PowerShell ;)
• @() has effect only on items that are not arrays. If you chain @(@(@(1))), you will get an array with one int item (@(@(@(1)))[0].gettype() returns Int32).
  So, @( ,@('r', 's') ) is the same as ,@('r', 's').

Answer 2

I tried what stej said and it worked, using this example:

function funC([int]$numOfPairs)
{
  $ret = @()
  if($numOfPairs -eq 1)
  { $ret = ,@('r','s') }
  elseif($numOfPairs -eq 2)
  { $ret = @('r','s'),@('t','u') }
  else
  { $ret = @('r','s'),@('t','u'),@('v','w') }

  Display $ret 0 "Inside Function C ($numOfPairs)"
  return ,$ret
}


### Start Program

$z = funC 1
Display $z 0 'Return from Function C(1)'
$z = funC 2
Display $z 0 'Return from Function C(2)'
$z = funC 3
Display $z 0 'Return from Function C(3)'