$_.Exception is $null when used in a call – why?

I have a PowerShell script containing the following logging function:

function LogError([string] $message, $exception = $null) {…}

In a try-catch block, when an exception occurs, I call that logging function like this:

catch { LogError("…", $_.Exception) }

In the LogError function, the second argument is always $null. Why?

I couldn’t find any documentation that would explain why I cannot use $_.Exception in a function call or that I am supposed to use instead.

>Solution :

Powershell function arguments are not passed using the parenthesis / comma formatting.

That is bad

LogError("…", $_.Exception)

Powershell take that as a single array argument.
This is actually the same as LogError -Message ("...",$_.Exception)

That is ok

LogError '…' $_.Exception

That is best

 LogError -message '...' -exception $_.Exception

Complete working example

function LogError([string] $message, $exception = $null) {
  Write-Host $message
   $exception | Out-String | Write-Host -ForegroundColor red 

try {
  throw 'Noooo !!!!'
catch {
  LogError -message 'oops' -exception $_.Exception

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