PowerShell pass by reference not working for me

Question

I'm trying to make a simple swap function in PowerShell, but passing by reference doesn't seem to work for me.

function swap ([ref]$object1, [ref]$object2){
  $tmp = $object1.value
  $object1.value = $object2.value
  $object2.value = $tmp
}

$a = 1
$b = 2
$a, $b
swap ([ref]$a) ,([ref]$b)
$a, $b

This SHOULD work, but no...

Output:
    1
    2
    1
    2

What did I do wrong?

score 29 · Accepted Answer · edited May 23 '17 at 11:46

29

Call like this:

swap ([ref]$a) ([ref]$b)

The mistake of using , is described in the Common Gotchas for PowerShell here on Stack Overflow.

edited May 23 '17 at 11:46

Community

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answered Aug 22 '11 at 15:40

manojlds

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score 25 · Answer 2 · edited Dec 28 '18 at 13:16

25

By the way, PowerShell has a special syntax to swap values, and there isn't a need to use $tmp:

$a,$b = $b,$a

edited Dec 28 '18 at 13:16

Peter Mortensen

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answered Aug 22 '11 at 15:45

Shay Levy

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score 3 · Answer 3 · edited Dec 28 '18 at 13:15

3

Firstly, you're calling it wrong. Putting a comma in the call to swap means you're passing an array of them to objects as the first parameter. If you were to correct it...

swap ([ref]$a) ([ref]$b)

...it would then work.

edited Dec 28 '18 at 13:15

Peter Mortensen

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answered Aug 22 '11 at 15:38

Joel B Fant

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You can pass value types by reference. – Rynant Aug 22 '11 at 16:12
@Rynant: You're right. I did it over again and it didn't complain about value types. Not sure what I did different the first time. – Joel B Fant Aug 22 '11 at 16:20

PowerShell pass by reference not working for me

3 Answers3

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