New closure on scriptblock

问题:

Consider this code:

PS> $timer = New-Object Timers.Timer
PS> $timer.Interval = 1000
PS> $i = 1;
PS> Register-ObjectEvent $timer Elapsed -Action { write-host 'i: ' $i }.GetNewClosure()
PS> $timer.Enabled = 1
i:  1
i:  1
i:  1
 ...
# wait a couple of seconds and change $i
PS> $i = 2
i:  2
i:  2
i:  2

I assumed that when I create new closure ({ write-host 'i: ' $i }.GetNewClosure()) value of $i will be tied to this closure. But not in this case. Afer I change the value, write-host takes the new value.

On the other side, this works:

PS> $i = 1;
PS> $action = { write-host 'i: ' $i }.GetNewClosure()
PS> &$action
i:  1
PS> $i = 2
PS> &$action
i:  1

Why it doesn't work with the Register-ObjectEvent?

回答1:

Jobs are executed in a dynamic module; modules have isolated sessionstate, and share access to globals. PowerShell closures only work within the same sessionstate / scope chain. Annoying, yes.

-Oisin

p.s. I say "jobs" because event handlers are effectively local jobs, no different than script being run with start-job (local machine only, implicit, not using -computer localhost)

回答2:

I think you are making assumptions that don't hold. PSH is interpreted, so when a code block is created it just holds the source code. When it is later evaluated any variables it uses will be looked up in the normal PSH way: first in the current scope, and then in each outer scope until a variable with a matching name if found.

When the timer fires its event, it executes the code block and thus looks up $i. Which is found in the outer scope with a value of 2.

In the second case, if you just use the code block directly (remove call to GetNewClosure) then the second execution gives 2.

回答3:

Use global variables in that case:

PS> $global:i = 1
PS> $timer = New-Object Timers.Timer
PS> $timer.Interval = 1000
PS> Register-ObjectEvent $timer Elapsed -Action { write-host 'i: ' $global:i }.GetNewClosure()
PS> $timer.Enabled = 1
i:  1
i:  1
i:  1

PS> Set-Variable -Name i -Value 2 -Scope Global

i:  2
i:  2
i:  2

Source:

http://stackoverflow.com/q/12535419/1287856