Powershell Start-job invoke function argument with parameter

Question

I have a function mainFunction that gets 2 parameters - $name will be just a regular string, and $moveFunction will be some function.

I want to start a job of a ScriptBlock ($SB) that will invoke $moveFunction with $name as his argument.

function foo($a){
    Write-Output "In function foo with the argument => $a"
}

$SB = { 
        param($C, $fooFunction)
        $fooFunction.Invoke($C)
     } 

function mainFunction($name, $moveFunction){
    Start-Job  -Name "currentJob" -ArgumentList $name, ${Function:$moveFunction} -ScriptBlock $SB

}


$j1 = mainFunction -name "output!" -moveFunction $Function:foo

I checked that $moveFunction exists in mainFunction already ($moveFunction.invoke(5) at mainFunction)

I can't find the problem in passing the function as argument in the start-job.

and from Get-Job -Name "CurrentJob" | Receive-Job I get:

You cannot call a method on a null-valued expression.
    + CategoryInfo          : InvalidOperation: (:) [], RuntimeException
    + FullyQualifiedErrorId : InvokeMethodOnNull
    + PSComputerName        : localhost

Any help would be appreciated.

edit: The problem is most likely the way I pass the function as an argument (${Function:$moveFunction}

I did a little poking around and it looks like someone already answered this type of question [here](https://stackoverflow.com/a/11844992/5339918). Essentially, Args for Jobs and Remote commands are Serialized. ScriptBlock comes out the other end as a String. Just call `$func = [ScriptBlock]::Create($func)` to convert that string back into a SB proper and invoke as per normal. — RiverHeart, Nov 26 '18 at 01:43
@RiverHeart Eventually your comment really helped me. Add your answer I'll approve it =) — sheldonzy, Nov 27 '18 at 14:39

Prasoon Karunan V · Answer 1 · 2018-11-27T06:31:40.467

1

You passing the same function and invoking it. You can directly use the function in the job.

Start-Job  -Name "currentJob" -ArgumentList $name - ScriptBlock ${function:foo}

edited Nov 27 '18 at 06:31

answered Nov 26 '18 at 03:39

Prasoon Karunan V

2,916
2
12
26

I'll look at your answer at depth tomorrow, but from a fast run it doesn't go. The call is still `$j1 = mainFunction -name "output!" -moveFunction $Function:foo` ? – sheldonzy Nov 26 '18 at 08:42
You can even `Start-Job -Name "currentJob" -ArgumentList $name, $Function:foo - ScriptBlock $SB` – Prasoon Karunan V Nov 26 '18 at 12:39
Did nobody read my comment from above? :( This doesn't work because the script block is serialized and comes out as a string when passed to the job. You need to take that string and generate a new scriptblock. – RiverHeart Nov 26 '18 at 22:14
yup, It will de-serialize, I didn't notice your comment. – Prasoon Karunan V Nov 27 '18 at 06:30
@sheldonzyyou can directly call the function in the job. – Prasoon Karunan V Nov 27 '18 at 06:31

score 1 · Accepted Answer · answered Nov 27 '18 at 22:32

Just a rehash of my previous comment plus code example. Similar issue here. Essentially, arguments passed to Jobs and Remote commands are serialized. During the de-serialization process, functions and script blocks come out as strings instead of their original type. Fortunately it's a simple process to transform these into invokable scriptblocks using [ScriptBlock]::Create("string").

function foo {
    write-host "foo"
}

function bar {
    # This argument comes in as a string
    param($func)
    write-host "bar"

    # Create scriptblock from string
    $func = [ScriptBlock]::Create($func)
    $func.invoke()
}

Start-Job -ArgumentList $Function:Foo -ScriptBlock $Function:Bar

Get-Job | Wait-job
Get-Job | Receive-job

Powershell Start-job invoke function argument with parameter

2 Answers2

Linked