How does the compound assignment operator behave in child scope? - powershell

Question

What happens to the following two function? I expect they are the same.
PS C:\> $i = 5
PS C:\> function Test1 { $i += 1; $i }
PS C:\> function Test2 { $i = $i + 1; $i }
PS C:\> Test1
1 # why?
PS C:\> Test2
6
PS C:\> $i
5 # will not change
I'm aware the $i in the function has local scope, hence will not be changed in the global scope, this is intentional. This question is simply about why the following 2 assignment statements behave differently here, as far as I know, they should be equivalent.
$i = $i + 1
$i += 1

Answer 1

Explanation:
In Test1 the variable $i is assigned a value (using the compound assignment operator). Because you cannot change the variable from the global scope, a new local variable is created (hiding the global variable), which does not have a value initially (basically 0), which is then increased by 1. When returning the value, the local variable is used.
function Test1 {
# because it's an assignment, a new local variable
# is created (with initial value = 0)
$local:i += 1
$local:i
}
In Test2 the value of the global variable $i is used (because the variable is visible in this child scope), 1 is added, and the result (6) is assigned to a new local variable. As in Test1, the value of the local variable is returned at the end.
function Test2 {
# a new local variable is created, but because the global
# variable is visible here, its value is used initially
$local:i = $global:i + 1
$local:i
}
Read more about scopes here: about_scopes

Answer 2

As for ...
I want the behavior of function Test2
($i = 5)
# Results
<#
5
#>
<#
To save the time/code in calling the variable separately, you can use PowerShell
variable squeezing to assign to the variable and output to the screen at the same
time.
This is more direct if you are always going to output to the screen.
It's a choice and has no effect on what you are doing
function Test1 { ($i += 1) }
function Test2 { ($i = $i + 1) }
Test1
Test2
# Results
<#
1
6
#>
($i = 5)
# Note the small refactor in the operator
function Test1 { ($i + 1) }
function Test2 { ($i = $i + 1) }
Test1
Test2
# Results
<#
6
6
#>
This still is a scope issue, local vs global, but the above gives you the matching results in both calls.