Vista - Re: Leading comma operator - What does it do, really?

09-15-2009

Good question. I'm not aware of any definitive article.

An easy way to see the difference:
PS>(1,2)|get-member
vs
PS>,(1,2)|get-member

In the first case, each element is passed to get-member. In the second
case, the elements remain grouped together and are passed as a whole to
get-member.

So at first, PowerShell sees the numbers 1 and 2 being passed, while in the
second, PowerShell sees a collection of 1 *and* 2 being passed together.

That being said, I'm not sure what's going on in your last example below.

Marco

"Kevin Buchan" <kevin.buchan@newsgroup[PlsDon'tSpam]sanders.com> wrote in
message news:n1bva5173gsjipm0gbsdimbiahafkv1q78@newsgroup

Quote:

> I'm having trouble finding a definitive article on what exactly the
> leading comma does.
>
> PS C:\> $a
> PS C:\> $a = 1,2,3
> PS C:\> $b = 4,5,6
> PS C:\> ($a + $b).length
> 6
>
> It *seems* to treat what follows it as an element in an array, even if
> it's an array. So instead of creating a new array with all of the
> elements in $a and $b, it creates an array with all of the elements in
> $a and another array element with the entire $b array as that single
> element.
>
> PS C:\> ($a + ,$b).length
> 4
> PS C:\>
> PS C:\> $c = $a + ,$b
> PS C:\> $c[3].GetType().FullName
> System.Object[]
> PS C:\> $c[3]
> 4
> 5
> 6
> PS C:\> $c[3][0]
> 4
>
> When I try it before a string, I get an array with a single element,
> the string.
> PS C:\> $d = ,"Test","AnotherTest"
> PS C:\> $d.GetType().FullName
> System.Object[]
> PS C:\> $d.length
> 2
> PS C:\> $d[0]
> Test
>
>
> OK, all of that seems reasonable to me. But, now here's where my mind
> gets blown...
>
> Why, in this example below, is the first element of $e a
> multi-dimensional array instead of just being a simple array of
> integers?
> PS C:\> $e = ,(1,2,3),4,5,6
> PS C:\> $e.length
> 4
> PS C:\> $e[0].GetType().FullName
> System.Object[]
> PS C:\> $e[0][1] # I expected this to return '2'.
> PS C:\> $e[0][0]
> 1
> 2
> 3
> PS C:\> $e[0][0][1] # What the heck?
> 2
>
>
> --
> Kevin Buchan
> kevin.buchan@newsgroup[nospam]sanders.com

09-15-2009

Ditto 'good question'.

To paraphrase your example

$e = ,(1,2,3),4,5,6

behaves in the same way as
$e = ,,(1,2,3) + 4,5,6

whereas one would expect it to be semantically equivalent to
$e = ,(1,2,3) + 4,5,6

Possibly a bug ?

--
Jon

"Kevin Buchan" <kevin.buchan@newsgroup[PlsDon'tSpam]sanders.com> wrote in
message news:n1bva5173gsjipm0gbsdimbiahafkv1q78@newsgroup

Quote:

> I'm having trouble finding a definitive article on what exactly the
> leading comma does.
>
> PS C:\> $a
> PS C:\> $a = 1,2,3
> PS C:\> $b = 4,5,6
> PS C:\> ($a + $b).length
> 6
>
> It *seems* to treat what follows it as an element in an array, even if
> it's an array. So instead of creating a new array with all of the
> elements in $a and $b, it creates an array with all of the elements in
> $a and another array element with the entire $b array as that single
> element.
>
> PS C:\> ($a + ,$b).length
> 4
> PS C:\>
> PS C:\> $c = $a + ,$b
> PS C:\> $c[3].GetType().FullName
> System.Object[]
> PS C:\> $c[3]
> 4
> 5
> 6
> PS C:\> $c[3][0]
> 4
>
> When I try it before a string, I get an array with a single element,
> the string.
> PS C:\> $d = ,"Test","AnotherTest"
> PS C:\> $d.GetType().FullName
> System.Object[]
> PS C:\> $d.length
> 2
> PS C:\> $d[0]
> Test
>
>
> OK, all of that seems reasonable to me. But, now here's where my mind
> gets blown...
>
> Why, in this example below, is the first element of $e a
> multi-dimensional array instead of just being a simple array of
> integers?
> PS C:\> $e = ,(1,2,3),4,5,6
> PS C:\> $e.length
> 4
> PS C:\> $e[0].GetType().FullName
> System.Object[]
> PS C:\> $e[0][1] # I expected this to return '2'.
> PS C:\> $e[0][0]
> 1
> 2
> 3
> PS C:\> $e[0][0][1] # What the heck?
> 2
>
>
> --
> Kevin Buchan
> kevin.buchan@newsgroup[nospam]sanders.com

09-15-2009

PS C:\> $e.SyncRoot

helps illuminate the structure of $e
Try, for example

PS C:> $e = ,(1,2),,(3,4),,(5,6)
PS C:> $e.SyncRoot

See also
http://devcentral.f5.com/weblogs/Joe...nraveling.aspx

- Larry

Kevin Buchan wrote:

Quote:

> ...
> Why, in this example below, is the first element of $e a
> multi-dimensional array instead of just being a simple array of
> integers?
> PS C:\> $e = ,(1,2,3),4,5,6
> PS C:\> $e.length
> 4
> PS C:\> $e[0].GetType().FullName
> System.Object[]
> PS C:\> $e[0][1] # I expected this to return '2'.
> PS C:\> $e[0][0]
> 1
> 2
> 3
> PS C:\> $e[0][0][1] # What the heck?
> 2
>

09-15-2009

PS C:> $e = ,(1,2),,(3,4),,(5,6)
can be written as
PS C:> $e = (,(1,2)),(,(3,4)),(,(5,6))

- Larry

Larry__Weiss wrote:

Quote:

> PS C:\> $e.SyncRoot
>
> helps illuminate the structure of $e
> Try, for example
>
> PS C:> $e = ,(1,2),,(3,4),,(5,6)
> PS C:> $e.SyncRoot
>
> See also
> http://devcentral.f5.com/weblogs/Joe...nraveling.aspx
>
> Kevin Buchan wrote:

Quote:

> > ...
>> Why, in this example below, is the first element of $e a
>> multi-dimensional array instead of just being a simple array of
>> integers?
>> PS C:\> $e = ,(1,2,3),4,5,6
>> PS C:\> $e.length
>> 4
>> PS C:\> $e[0].GetType().FullName
>> System.Object[]
>> PS C:\> $e[0][1] # I expected this to return '2'.
>> PS C:\> $e[0][0]
>> 1
>> 2
>> 3
>> PS C:\> $e[0][0][1] # What the heck?
>> 2
>>

09-15-2009	#1 (permalink)
Marco Shaw [MVP] n/a posts	Re: Leading comma operator - What does it do, really? Good question. I'm not aware of any definitive article. An easy way to see the difference: PS>(1,2)\|get-member vs PS>,(1,2)\|get-member In the first case, each element is passed to get-member. In the second case, the elements remain grouped together and are passed as a whole to get-member. So at first, PowerShell sees the numbers 1 and 2 being passed, while in the second, PowerShell sees a collection of 1 and 2 being passed together. That being said, I'm not sure what's going on in your last example below. Marco "Kevin Buchan" <kevin.buchan@newsgroup[PlsDon'tSpam]sanders.com> wrote in message news:n1bva5173gsjipm0gbsdimbiahafkv1q78@newsgroup Quote: > I'm having trouble finding a definitive article on what exactly the > leading comma does. > > PS C:\> $a > PS C:\> $a = 1,2,3 > PS C:\> $b = 4,5,6 > PS C:\> ($a + $b).length > 6 > > It seems to treat what follows it as an element in an array, even if > it's an array. So instead of creating a new array with all of the > elements in $a and $b, it creates an array with all of the elements in > $a and another array element with the entire $b array as that single > element. > > PS C:\> ($a + ,$b).length > 4 > PS C:\> > PS C:\> $c = $a + ,$b > PS C:\> $c[3].GetType().FullName > System.Object[] > PS C:\> $c[3] > 4 > 5 > 6 > PS C:\> $c[3][0] > 4 > > When I try it before a string, I get an array with a single element, > the string. > PS C:\> $d = ,"Test","AnotherTest" > PS C:\> $d.GetType().FullName > System.Object[] > PS C:\> $d.length > 2 > PS C:\> $d[0] > Test > > > OK, all of that seems reasonable to me. But, now here's where my mind > gets blown... > > Why, in this example below, is the first element of $e a > multi-dimensional array instead of just being a simple array of > integers? > PS C:\> $e = ,(1,2,3),4,5,6 > PS C:\> $e.length > 4 > PS C:\> $e[0].GetType().FullName > System.Object[] > PS C:\> $e[0][1] # I expected this to return '2'. > PS C:\> $e[0][0] > 1 > 2 > 3 > PS C:\> $e[0][0][1] # What the heck? > 2 > > > -- > Kevin Buchan > kevin.buchan@newsgroup[nospam]sanders.com
My System Specs

09-15-2009	#2 (permalink)
Jon n/a posts	Re: Leading comma operator - What does it do, really? Ditto 'good question'. To paraphrase your example $e = ,(1,2,3),4,5,6 behaves in the same way as $e = ,,(1,2,3) + 4,5,6 whereas one would expect it to be semantically equivalent to $e = ,(1,2,3) + 4,5,6 Possibly a bug ? -- Jon "Kevin Buchan" <kevin.buchan@newsgroup[PlsDon'tSpam]sanders.com> wrote in message news:n1bva5173gsjipm0gbsdimbiahafkv1q78@newsgroup Quote: > I'm having trouble finding a definitive article on what exactly the > leading comma does. > > PS C:\> $a > PS C:\> $a = 1,2,3 > PS C:\> $b = 4,5,6 > PS C:\> ($a + $b).length > 6 > > It seems to treat what follows it as an element in an array, even if > it's an array. So instead of creating a new array with all of the > elements in $a and $b, it creates an array with all of the elements in > $a and another array element with the entire $b array as that single > element. > > PS C:\> ($a + ,$b).length > 4 > PS C:\> > PS C:\> $c = $a + ,$b > PS C:\> $c[3].GetType().FullName > System.Object[] > PS C:\> $c[3] > 4 > 5 > 6 > PS C:\> $c[3][0] > 4 > > When I try it before a string, I get an array with a single element, > the string. > PS C:\> $d = ,"Test","AnotherTest" > PS C:\> $d.GetType().FullName > System.Object[] > PS C:\> $d.length > 2 > PS C:\> $d[0] > Test > > > OK, all of that seems reasonable to me. But, now here's where my mind > gets blown... > > Why, in this example below, is the first element of $e a > multi-dimensional array instead of just being a simple array of > integers? > PS C:\> $e = ,(1,2,3),4,5,6 > PS C:\> $e.length > 4 > PS C:\> $e[0].GetType().FullName > System.Object[] > PS C:\> $e[0][1] # I expected this to return '2'. > PS C:\> $e[0][0] > 1 > 2 > 3 > PS C:\> $e[0][0][1] # What the heck? > 2 > > > -- > Kevin Buchan > kevin.buchan@newsgroup[nospam]sanders.com
My System Specs

09-15-2009	#3 (permalink)
Larry__Weiss n/a posts	Re: Leading comma operator - What does it do, really? PS C:\> $e.SyncRoot helps illuminate the structure of $e Try, for example PS C:> $e = ,(1,2),,(3,4),,(5,6) PS C:> $e.SyncRoot See also http://devcentral.f5.com/weblogs/Joe...nraveling.aspx - Larry Kevin Buchan wrote: Quote: > ... > Why, in this example below, is the first element of $e a > multi-dimensional array instead of just being a simple array of > integers? > PS C:\> $e = ,(1,2,3),4,5,6 > PS C:\> $e.length > 4 > PS C:\> $e[0].GetType().FullName > System.Object[] > PS C:\> $e[0][1] # I expected this to return '2'. > PS C:\> $e[0][0] > 1 > 2 > 3 > PS C:\> $e[0][0][1] # What the heck? > 2 >
My System Specs

09-15-2009	#4 (permalink)
Larry__Weiss n/a posts	Re: Leading comma operator - What does it do, really? PS C:> $e = ,(1,2),,(3,4),,(5,6) can be written as PS C:> $e = (,(1,2)),(,(3,4)),(,(5,6)) - Larry Larry__Weiss wrote: Quote: > PS C:\> $e.SyncRoot > > helps illuminate the structure of $e > Try, for example > > PS C:> $e = ,(1,2),,(3,4),,(5,6) > PS C:> $e.SyncRoot > > See also > http://devcentral.f5.com/weblogs/Joe...nraveling.aspx > > Kevin Buchan wrote: Quote: > > ... >> Why, in this example below, is the first element of $e a >> multi-dimensional array instead of just being a simple array of >> integers? >> PS C:\> $e = ,(1,2,3),4,5,6 >> PS C:\> $e.length >> 4 >> PS C:\> $e[0].GetType().FullName >> System.Object[] >> PS C:\> $e[0][1] # I expected this to return '2'. >> PS C:\> $e[0][0] >> 1 >> 2 >> 3 >> PS C:\> $e[0][0][1] # What the heck? >> 2 >>
My System Specs

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