Question

The following powershell script

"one","two","three" | % { "$(($l++)): $_" }

will print

1: one
2: two
3: three

However, after remove the bracket around $l++

"one","two","three" | % { "$($l++): $_" }

it will print

: one
: two
: three

Answer 1

This is because $l++ is a voidable statement. In powershell certain types of
expressions, when used as statements, are not displayed.
Voidable statements include assignments and the increment/decrement operators. When they are used in an expression, they return a value, but when they’re used as a standalone statement, they return no value. It is very well explained in Windows Powershell in Action by Bruce Payette:

The increment and decrement operators were almost not included in PowerShell
because they introduced a problem. In languages such as C and C#, when you use
one of these operators as a statement:
$a++
nothing is displayed. This is because statements in C and C# don’t return values. In
PowerShell, however, all statements return a value. This led to confusion. People
would write scripts like this:

$sum=0
$i=0
while ($i -lt 10) { $sum += $i; $i++ }
$sum

and be surprised to see the numbers 1 through 10 displayed. This was because $a++
returned a value and PowerShell was displaying the results of every statement. This
was so confusing that we almost removed these operators from the language. Then we
hit on the idea of a voidable statement. Basically, this means that certain types of
expressions, when used as statements, are not displayed. Voidable statements include
assignments and the increment/decrement operators. When they are used in an
expression, they return a value, but when they’re used as a standalone statement, they
return no value. Again, this is one of those details that won’t affect how you use PowerShell other than to make it work as you expect. (source: Windows Powershell in Action)