Vista - Subexpressions and Unary Operators

Hi

I'm trying to work out the difference in the behaviour of $() and () - in
particular with regard to unary operators, and would appreciate any insights
on the matter.


#These statements work as expected ....

$i=3; Write-Host ($i++)
$i=3; Write-Host (++$i)

#but these show no output

$i=3; Write-Host $($i++)
$i=3; Write-Host $(++$i)
$i=3; $i++
$i=3; ++$i


#whereas these do ...

$i=3; Write-Host $(($i++))
$i=3; Write-Host $((++$i))


Any thoughts?

--
Jon

"Jon" <Email_Address@xxxxxx> wrote in message
news:OvU36gLfIHA.2000@xxxxxxBruce Payette talks about this in his book Windows PowerShell in Action. It
basically comes down to PowerShell making assumptions about whether or not
you want to see the result of an increment/decrement operation. Most of the
time you don't. However () is special since you can only have one
expression in it and if they followed their normal rules then ($i++)
wouldn't output anything. The language designers figured that would be
"unexpected" in this scenario so they make an exception.

HTH,
Keith

"Keith Hill [MVP]" <r_keith_hill@xxxxxx_spam_I> wrote in message
news:E71EA756-D026-4A25-A642-4F9096F4D22A@xxxxxx

Thanks Keith - one of the books on my "to read " list. It sounds like it's
been reasoned through then.

It's slightly different from what I've experienced in other languages, hence
the question. I suppose as it stands now, the flexibility is there to either
output or not output as you choose. Hopefully that flexibility will remain,
since it's a useful construction with commands like Write-Host - otherwise
you're forced to use roundabout techniques like

$var = 3; $var = $var + 1
Write-Host "Var is now $var"

#More compact, as you can now, to be able to write
$var = 3
Write-Host "Var is now $((++$var))"

albeit a little unintuitive.

--
Jon

"Jon" <Email_Address@xxxxxx> wrote in message
news:u5Dew#LfIHA.4712@xxxxxxOr you can do this:

$var = 0
Write-Host "$($var++;$var)"

which is bit easier to understand.

--
Keith