PowerShell get out result from a function - powershell

Question

I have the following code that I'm trying to return the string Good as an out parameter from FooFunc.
How can I do that?
function FooFunc($a, [ref]$result){
if (4 -gt 1) {
$result = "Good"
return $true
} else {
$result = "Bad"
return $false
}
}
try
{
FooFunc "Bar" ([ref]$result)
Write-Host $result
}
catch
{
Write-Host $_.Exception.Message
}
EDIT I don't want to return the answer with return, I need it as ref or out parameters

Answer 1

You can try
Function Foo{
if(4 -gt 1)
{
return "Good"
}
else
{
return "Bad"
}
}
try{
$result = Foo
Write-Output $result
}
catch{
Write-Output "Some message"
}

Answer 2

[ref] is for passing variable into a function so it needs to exist in your current scope which is why you are getting the error. I think you are misunderstanding the usage. Try this:
function FooFunc([ref]$result){
if (4 -gt 1) {
$result = 'good' + $result.value
$result
} else {
$result = "Bad"
return $false
}
}
try
{
$result = 'badval'
FooFunc ([ref]$result)
}
catch
{
Write-Host $_.Exception.Message
}
If you pipe the function to get-member you will see it returns ref and the output is using the variable you are passing in.

Answer 3

Applying [ref] to a parameter will result in a PSReference object being passed to the function.
In the case of function FooFunc($a, [ref]$result), using $result = "Good" will change the local $result variable in the function from a PSReference to a string, but will have no effect on the variable the caller passed in. To modify the outer variable, you would want to set $result.Value = "Good".
And as Noah Sparks mentioned, on the caller side you also need to declare the variable before using it as argument.
function FooFunc( $a, [ref]$result ) {
if( 4 -gt 1 ) {
$result.Value = 'Good' # change the reference value, not the variable
return $true
}
else {
$result.Value = 'Bad'
return $false
}
}
try {
$result = $null # declare variable before using it for a reference
FooFunc 'Bar' ([ref]$result)
Write-Host $result
}
catch {
Write-Host $_.Exception.Message
}
Another option would be to output multiple values from your function:
function FooFunc( $a ) {
if( 4 -gt 1 ) {
# Output both result and success values
'Good'
return $true
}
else {
# This single return is the same as the output and return above
return 'Bad', $false
}
}
try {
$result,$success = FooFunc 'Bar' # Receive both values
$success # Implicitly output success value, as before
Write-Host $result
}
catch {
Write-Host $_.Exception.Message
}