When does the start-process get executed?

Question

Basically I'm trying to return exit codes from start-process to the script so that MDT/SCCM can fail correctly if an installation fails.

Generically here is the code I'm using:

$proc = Start-Process -FilePath $setupexe -ArgumentList $setupargs -Wait -Passthrough
Exit $proc.ExitCode

My question is when does the Start-Process get executed? When I define $proc or when I call $proc.ExitCode?

What I'm trying to do is use the exit code in a if statement without having to store that code in another variable (reduce code clutter).

$proc = Start-Process -FilePath $setupexe -ArgumentList $setupargs -Wait -PassThru
if ($proc.ExitCode -ne 0) {Exit $proc.ExitCode}

$proc2 = Start-Process -FilePath $setupexe2 -ArgumentList $setupargs2 -Wait -PassThru
if ($proc2.ExitCode -ne 0) {Exit $proc.ExitCode}

vs

$proc = Start-Process -FilePath $setupexe -ArgumentList $setupargs -Wait -PassThru
$procexit = $proc.ExitCode
if ($procexit -ne 0) {Exit $procexit}

$proc2 = Start-Process -FilePath $setupexe2 -ArgumentList $setupargs2 -Wait -PassThru
$procexit2 - $proc2.ExitCode
if ($procexit2 -ne 0) {Exit $procexit2}

I don't want the Start-Process to be called again just to kill the script and return the error code.

BenH · Accepted Answer · 2018-01-05T19:22:02.930

Start-Process will start the process when you define $proc and won't move on to the next line until it exits since you have the -Wait parameter defined.

This line if ($proc.ExitCode -ne 0) {Exit $proc.ExitCode} will not cause the code to run again.

You could test this on your computer by running the code with a quick program like notepad and see when program appears.

$a = start-process notepad.exe -wait -passthru
$a.exitcode

When does the start-process get executed?

1 Answers1