Powershell: Get object value to int array

Question

I'm fairly new to PowerShell

Can someone please explain the most simple way to copy values from an object into a basic integer array without also copying the original object.

Sorry if this is confusing, but what I mean is that I want to capture object values and put them in an integer array where I can then do simple arithmetic on the values

For example

$ComputerCpu = Get-WmiObject win32_processor -computer "MyComputer" | `
    Select-Object {$_.LoadPercentage}

On a Windows Server 2008 VM at work this returns two values

I want to get an average so I tried

$average = ($ComputerCpu[0] + $ComputerCpu[1]) / 2

This cannot work as the values are not of type Int.

I tried casting the values

$average = ([int]$ComputerCpu[0] + [int]$ComputerCpu[1]) / 2

But this results in a cannot convert error.

I know what the problem is but it would be great if someone could explain the fundamental solution to this as I run into this problem a lot and I know there must be a defined solution or a way to avoid this entirely.

Thanks.

PS: If possible please don't provide an alternative solution, as I really want to know how to get an objects values into an integer array so as to perform arithmetic on those values.

Thanks.

Select-Object -> ForEach-Object

user4003407
– user4003407

2016-02-29 18:46:14 +00:00
Commented Feb 29, 2016 at 18:46 — user4003407
– user4003407, Commented Feb 29, 2016 at 18:46

briantist · Accepted Answer · 2016-02-29 18:46:28Z

1

In addition to EBGreen's answer, you could use the -ExpandProperty parameter of Select-Object:

$ComputerCpu = Get-WmiObject win32_processor -computer "MyComputer" |
    Select-Object -ExpandProperty LoadPercentage

The rest of your code would work then.

answered Feb 29, 2016 at 18:46

briantist

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1 Comment

eodeluga Over a year ago

Thank you. This works where @EBGreen solution did not (but I don't know why).

EBGreen · Accepted Answer · 2016-02-29 18:46:48Z

1

Reference the property not the entire object. So:

$average = ($ComputerCpu[0].LoadPercentage + $ComputerCpu[1].LoadPercentage) / 2

answered Feb 29, 2016 at 18:46

EBGreen

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2 Comments

eodeluga Over a year ago

For some reason this returns $average as 0

briantist Over a year ago

@ATtheincredibleaf possibly the average was zero at the time (or less than 1 and truncated to [int])?

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