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Comma Before Parentheses in Variable Assignment

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2

1

I'm using the S.DS.P PowerShell module in a PowerShell script of mine. In it, I have to create the following object:

$ADUserEntry = @"distinguishedName"=$null;"objectClass"=$null;"sAMAccountName"=$null;"unicodePwd"=$null;"userAccountControl"=0;

In the documentation of the module, it's stated that I have to do the following assignment to the unicodePwd field of a variable created using this object:

$obj.unicodePwd = ,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]]);

Notice how there's a comma before the first parentheses. What is that comma doing there?

powershell

share|improve this question

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  it's the comma operator. [grin] it makes an array with the item to the left as an object in the new array. that can be used to prevent the automatic unrolling of an array by wrapping the 1st array in a 2nd one ... and the 2nd one is the one that gets unwrapped. you can see the MS docs on that here ... about_Operators | Microsoft Docs — docs.microsoft.com/en-us/powershell/module/…
  
  – Lee_Dailey
  Mar 26 at 20:59
  

  
  
  
  

add a comment | 

2

1

I'm using the S.DS.P PowerShell module in a PowerShell script of mine. In it, I have to create the following object:

$ADUserEntry = @"distinguishedName"=$null;"objectClass"=$null;"sAMAccountName"=$null;"unicodePwd"=$null;"userAccountControl"=0;

In the documentation of the module, it's stated that I have to do the following assignment to the unicodePwd field of a variable created using this object:

$obj.unicodePwd = ,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]]);

Notice how there's a comma before the first parentheses. What is that comma doing there?

powershell

share|improve this question

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  
  it's the comma operator. [grin] it makes an array with the item to the left as an object in the new array. that can be used to prevent the automatic unrolling of an array by wrapping the 1st array in a 2nd one ... and the 2nd one is the one that gets unwrapped. you can see the MS docs on that here ... about_Operators | Microsoft Docs — docs.microsoft.com/en-us/powershell/module/…
  
  – Lee_Dailey
  Mar 26 at 20:59
  

  
  
  
  

add a comment | 

I'm using the S.DS.P PowerShell module in a PowerShell script of mine. In it, I have to create the following object:

$ADUserEntry = @"distinguishedName"=$null;"objectClass"=$null;"sAMAccountName"=$null;"unicodePwd"=$null;"userAccountControl"=0;

In the documentation of the module, it's stated that I have to do the following assignment to the unicodePwd field of a variable created using this object:

$obj.unicodePwd = ,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]]);

Notice how there's a comma before the first parentheses. What is that comma doing there?

powershell

share|improve this question

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

$ADUserEntry = @"distinguishedName"=$null;"objectClass"=$null;"sAMAccountName"=$null;"unicodePwd"=$null;"userAccountControl"=0;

$obj.unicodePwd = ,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]]);

share|improve this question

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

share|improve this question

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

asked Mar 26 at 20:32

Gonçalo LourençoGonçalo Lourenço

77511 gold badge55 silver badges1717 bronze badges

1 Answer
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1

As Lee_Daily has pointed out, what you're seeing is the unary form of the (unfortunately named) "comma operator", i.e., PowerShell's array-construction operator.

The unary form creates a single-element array that wraps its (one and only) operand; the array's type is [object[]], as usual in PowerShell:

$arr = , 'foo' # wrap string 'foo' in a single-element array
$arr.GetType().Name # the array's type -> 'Object[]'
$arr[0].GetType().Name # the type of the array's one and only element -> 'String'

Note that while you can even wrap arrays that way, PowerShell's operator-precedence rules require a literal array operand to be enclosed in (...):

# OK - wraps array 1, 2 in a single-element array.
$arr = , (1, 2) 

# !! DOES SOMETHING DIFFERENT:
# Creates a 2-element array whose 1st element is integer 1 wrapped in a
# single-element array
$arr = , 1, 2

The binary form constructs an array from the operands, as expected:

$arr = 1, 2, 3 # 3-element array whose elements are integers 1 and 2 and 3

---

As an aside, re the specific command shown:

,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]])

Neither , nor -as [byte[]] are needed in this scenario, because
[System.Text.Encoding]::Unicode.GetBytes() directly returns a [byte[]] array.

share|improve this answer

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  My pleasure, @GonçaloLourenço; glad to hear it was helpful.
  
  – mklement0
  Mar 26 at 21:55
  

  
  
  
  

add a comment | 

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1

As Lee_Daily has pointed out, what you're seeing is the unary form of the (unfortunately named) "comma operator", i.e., PowerShell's array-construction operator.

The unary form creates a single-element array that wraps its (one and only) operand; the array's type is [object[]], as usual in PowerShell:

$arr = , 'foo' # wrap string 'foo' in a single-element array
$arr.GetType().Name # the array's type -> 'Object[]'
$arr[0].GetType().Name # the type of the array's one and only element -> 'String'

Note that while you can even wrap arrays that way, PowerShell's operator-precedence rules require a literal array operand to be enclosed in (...):

# OK - wraps array 1, 2 in a single-element array.
$arr = , (1, 2) 

# !! DOES SOMETHING DIFFERENT:
# Creates a 2-element array whose 1st element is integer 1 wrapped in a
# single-element array
$arr = , 1, 2

The binary form constructs an array from the operands, as expected:

$arr = 1, 2, 3 # 3-element array whose elements are integers 1 and 2 and 3

---

As an aside, re the specific command shown:

,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]])

Neither , nor -as [byte[]] are needed in this scenario, because
[System.Text.Encoding]::Unicode.GetBytes() directly returns a [byte[]] array.

share|improve this answer

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  My pleasure, @GonçaloLourenço; glad to hear it was helpful.
  
  – mklement0
  Mar 26 at 21:55
  

  
  
  
  

add a comment | 

1

As Lee_Daily has pointed out, what you're seeing is the unary form of the (unfortunately named) "comma operator", i.e., PowerShell's array-construction operator.

The unary form creates a single-element array that wraps its (one and only) operand; the array's type is [object[]], as usual in PowerShell:

$arr = , 'foo' # wrap string 'foo' in a single-element array
$arr.GetType().Name # the array's type -> 'Object[]'
$arr[0].GetType().Name # the type of the array's one and only element -> 'String'

Note that while you can even wrap arrays that way, PowerShell's operator-precedence rules require a literal array operand to be enclosed in (...):

# OK - wraps array 1, 2 in a single-element array.
$arr = , (1, 2) 

# !! DOES SOMETHING DIFFERENT:
# Creates a 2-element array whose 1st element is integer 1 wrapped in a
# single-element array
$arr = , 1, 2

The binary form constructs an array from the operands, as expected:

$arr = 1, 2, 3 # 3-element array whose elements are integers 1 and 2 and 3

---

As an aside, re the specific command shown:

,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]])

Neither , nor -as [byte[]] are needed in this scenario, because
[System.Text.Encoding]::Unicode.GetBytes() directly returns a [byte[]] array.

share|improve this answer

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  My pleasure, @GonçaloLourenço; glad to hear it was helpful.
  
  – mklement0
  Mar 26 at 21:55
  

  
  
  
  

add a comment | 

As Lee_Daily has pointed out, what you're seeing is the unary form of the (unfortunately named) "comma operator", i.e., PowerShell's array-construction operator.

The unary form creates a single-element array that wraps its (one and only) operand; the array's type is [object[]], as usual in PowerShell:

$arr = , 'foo' # wrap string 'foo' in a single-element array
$arr.GetType().Name # the array's type -> 'Object[]'
$arr[0].GetType().Name # the type of the array's one and only element -> 'String'

Note that while you can even wrap arrays that way, PowerShell's operator-precedence rules require a literal array operand to be enclosed in (...):

# OK - wraps array 1, 2 in a single-element array.
$arr = , (1, 2) 

# !! DOES SOMETHING DIFFERENT:
# Creates a 2-element array whose 1st element is integer 1 wrapped in a
# single-element array
$arr = , 1, 2

The binary form constructs an array from the operands, as expected:

$arr = 1, 2, 3 # 3-element array whose elements are integers 1 and 2 and 3

---

As an aside, re the specific command shown:

,([System.Text.Encoding]::Unicode.GetBytes("$randomPassword") -as [byte[]])

Neither , nor -as [byte[]] are needed in this scenario, because
[System.Text.Encoding]::Unicode.GetBytes() directly returns a [byte[]] array.

share|improve this answer

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges

$arr = , 'foo' # wrap string 'foo' in a single-element array
$arr.GetType().Name # the array's type -> 'Object[]'
$arr[0].GetType().Name # the type of the array's one and only element -> 'String'

# OK - wraps array 1, 2 in a single-element array.
$arr = , (1, 2) 

# !! DOES SOMETHING DIFFERENT:
# Creates a 2-element array whose 1st element is integer 1 wrapped in a
# single-element array
$arr = , 1, 2

$arr = 1, 2, 3 # 3-element array whose elements are integers 1 and 2 and 3

share|improve this answer

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges

answered Mar 26 at 21:21

mklement0mklement0

151k2525 gold badges272272 silver badges312312 bronze badges