Cannot assign value to variable from Powershell form - powershell

Question

Using the Microsoft Technet example here and pasted below
https://technet.microsoft.com/en-us/library/ff730941.aspx?f=255&MSPPError=-2147217396
I cannot get it to actually assign the user input to the variable $x
I've tried removing the void from this line at the end [void] $objForm.ShowDialog() but it appears to always return the Cancel text
[void] [System.Reflection.Assembly]::LoadWithPartialName("System.Drawing")
[void] [System.Reflection.Assembly]::LoadWithPartialName("System.Windows.Forms")
$objForm = New-Object System.Windows.Forms.Form
$objForm.Text = "Data Entry Form"
$objForm.Size = New-Object System.Drawing.Size(300,200)
$objForm.StartPosition = "CenterScreen"
$objForm.KeyPreview = $True
$objForm.Add_KeyDown({if ($_.KeyCode -eq "Enter")
{$x=$objTextBox.Text;$objForm.Close()}})
$objForm.Add_KeyDown({if ($_.KeyCode -eq "Escape")
{$objForm.Close()}})
$OKButton = New-Object System.Windows.Forms.Button
$OKButton.Location = New-Object System.Drawing.Size(75,120)
$OKButton.Size = New-Object System.Drawing.Size(75,23)
$OKButton.Text = "OK"
$OKButton.Add_Click({$x=$objTextBox.Text;$objForm.Close()})
$objForm.Controls.Add($OKButton)
$CancelButton = New-Object System.Windows.Forms.Button
$CancelButton.Location = New-Object System.Drawing.Size(150,120)
$CancelButton.Size = New-Object System.Drawing.Size(75,23)
$CancelButton.Text = "Cancel"
$CancelButton.Add_Click({$objForm.Close()})
$objForm.Controls.Add($CancelButton)
$objLabel = New-Object System.Windows.Forms.Label
$objLabel.Location = New-Object System.Drawing.Size(10,20)
$objLabel.Size = New-Object System.Drawing.Size(280,20)
$objLabel.Text = "Please enter the information in the space below:"
$objForm.Controls.Add($objLabel)
$objTextBox = New-Object System.Windows.Forms.TextBox
$objTextBox.Location = New-Object System.Drawing.Size(10,40)
$objTextBox.Size = New-Object System.Drawing.Size(260,20)
$objForm.Controls.Add($objTextBox)
$objForm.Topmost = $True
$objForm.Add_Shown({$objForm.Activate()})
[void] $objForm.ShowDialog()
$x

Answer 1

This code worked fine in PowerShell 1 and 2, but stopped working in PowerShell 3. PowerShell 3 offered a small set of syntax improvements and the usual new cmdlets, but on a less publicized note, it was rewritten on top of the Dynamic Language Runtime, a natural choice for a scripting language.
The move to the DLR changed the scoping rules in a way I don't think is well documented even to this day. Specifically, in a script block (like the one used to create a delegate) assigning to a variable creates a new variable in the scope of the block. This makes sense from the point of view of modularity, but isn't so obvious in a scripting language. In particular, consider the following:
$x = 2
&{
Write-Host $x;
$x = 3; Write-Host $x;
Remove-Variable "x"; Write-Host $x
$x = 3;
}
Write-Host $x
This prints:
2
3
2
2
Within the block, x takes its value from the parent scope, but when we assign it, we're actually creating a new variable in the private scope. When we remove that variable, any references to x are from the parent scope again. The assignment to x does not affect the value in the parent scope, as we can see from the Write-Host outside the block.
This explains why assigning a variable in a delegate "doesn't work": you're actually assigning to a new variable inside the delegate, the value of which is lost when the delegate ends. To assign to the variable in the parent scope, an explicit qualifier is necessary ($script:x or $global:x).