Is there a simple way to pass specific *named* PowerShell parameters through directly to a called function?

Question

I am sure I read somewhere that there is an easy way to pass named parameters from a calling function to a called function without explicitly naming and specifying each parameter.

This is more than just reusing the position; I'm interested in the case where the name of the passed parameters is the same in some cases, but not in others.

I also think there is a way that is not dependent on position.

function called-func {
    param([string]$foo, [string]$baz, [string]$bar)
    write-debug $baz
    write-host $foo,$bar
}

function calling-func {
    param([int]$rep = 1, [string]$foo, [string]$bar)
    1..$rep | %{
        called-func -foo $foo -bar $bar -baz $rep ## <---- Should this be simpler?
    }
}

calling-func -rep 10 -foo "Hello" -bar "World"

What would the method be, and is there a link?

I thought it might have been Jeffrey Snover, but I'm not sure.

Schuster

I did do some googling first, but since I didn't see the answer already here on SO, I wanted to get it here too anyway. – Schuster

So after some googling, I think the feature I was thinking of had to do with 'splatting' and passing switch parameters. I'll post an answer after I get all the details. – Schuster

Answer 1

In PowerShell v2 (which admittedly you may not be ready to move to yet) allows you to pass along parameters without knowing about them ahead of time:

called-func $PSBoundParameters

PSBoundParameters is a dictionary of all the parameters that were actually provided to your function. You can remove parameters you don't want (or add I suppose).

Icecap

Yes, I can't wait till v2 goes to production :) – Schuster

Answer 2

Well, I think I was confusing a blog post I read about switch parameters. As far as I can tell the best way is to just reuse the parameters like so:

called-func -foo:$foo -bar:$bar

Schuster

The colon syntax is required when foo is a switch parameter. If you pass -foo $foo then called-func will think it's receiving two arguments (the switch foo and a value $foo). – Cross

Answer 3

How about

called-func  $foo $bar

Corrode