Powershell array assignment assigns variable, not value?

Question

I have an example of a program that creates an array, and then attempts to assign the value of that array multiple times into another array as a multidimensional array.

$a =@(0,0,0)
$b = @($a,$a,$a)
$b[1][2]=2
$b
'And $a is changed too:'
$a

The output is:

PS E:\Workarea> .\what.ps1
0
0
2
0
0
2
0
0
2
And $a is changed too:
0
0
2

So in this instance, the variable is actually pointing to the original variable. This is very unexpected behavior. It's rather neat that one can do this, although I never did use unions that much in my C programming. But I'd like a way to actually just do the assignment of the value, not of the variable.

$b = @($a.clone(),$a.clone(),$a.clone())

I guess would work, but something tells me that there may be something a little more elegant than that.

Thanks for the input.

This is PowerShell 2.0 under Windows 7 64-bit.

Answer 1

To assign the values of $a to $b instead of the reference to $a, you can wrap the variable in $(). Anything in $() gets evaluated before using it in the command, so $($a) is equivalent to 0, 0, 0.

$a =@(0,0,0)
$b = @($($a),$($a),$($a))
$b[1][2]=2
$b
'$a is not changed.'
$a

Answer 2

Be careful in PowerShell ',' is not the enumerator operator, but an array aoperator. The thing you present as multidimensional array is in fact an array of array, you'll find here under the definition of a multidimensional array :

$a= new-object ‘object[,]’ 3,3
$a[0,2]=3 
PS > for ($i=0;$i -lt 3;$i++)
>> {
>> for($j=0;$j -lt 3;$j++)
>> {
>> $a[$i,$j]=$i+$j
>> }
>> }

Everything work here as $b is an array of reference.