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I see this when debubbing a filter definition like:
filter foo {
"`$_ is $_"
}
In the Variables window $_ shows up with no value. I believe the problem is that the code that checks to see if the variable is expandable returns true on primitives wrapped in a PSObject. I have a fix I'll submit in a few.
The text was updated successfully, but these errors were encountered:
rkeithhill
changed the title
Variables in the debugger are not displayed is type is PSObject
Variables in the debugger are not displayed if type is PSObject
Dec 7, 2015
I see this when debubbing a filter definition like:
In the Variables window
$_
shows up with no value. I believe the problem is that the code that checks to see if the variable is expandable returns true on primitives wrapped in a PSObject. I have a fix I'll submit in a few.The text was updated successfully, but these errors were encountered: