Issue with using invoke-expression on a char

Trying to add 2 different numbers together using something like:
invoke-expression 3${a}4

What I'd really like to have working though is something like this:
invoke-expression 3${a[0]}4

Now, $a.gettype() and $a[0].gettype() come out differently, and I don't know
if/how to make $a[0] into a string like $a would be.

[C:\psh]
121> $a="+"
[C:\psh]
122> $a.gettype()

IsPublic IsSerial Name BaseType
-------- -------- ---- --------
True True String System.Object

[C:\psh]
123> $a[0].gettype()

IsPublic IsSerial Name BaseType
-------- -------- ---- --------
True True Char System.ValueType

[C:\psh]
124> invoke-expression 3${a}4
7
[C:\psh]
125> invoke-expression 3${a[0]}4
34
[C:\psh]

Is this what you had in mind?

$op = "+"
invoke-expression "3 $op 3"
6

--
William Stacey [C# MVP]
PCR concurrency library: www.codeplex.com/pcr
PSH Scripts Project www.codeplex.com/psobject


"Marco Shaw" <marcoDOTshaw_@_gmailDOTcom> wrote in message
news:%237bG9yTUHHA.3592@TK2MSFTNGP03.phx.gbl...
| Trying to add 2 different numbers together using something like:
| invoke-expression 3${a}4
|
| What I'd really like to have working though is something like this:
| invoke-expression 3${a[0]}4
|
| Now, $a.gettype() and $a[0].gettype() come out differently, and I don't
know
| if/how to make $a[0] into a string like $a would be.
|
| [C:\psh]
| 121> $a="+"
| [C:\psh]
| 122> $a.gettype()
|
| IsPublic IsSerial Name BaseType
| -------- -------- ---- --------
| True True String System.Object
|
| [C:\psh]
| 123> $a[0].gettype()
|
| IsPublic IsSerial Name BaseType
| -------- -------- ---- --------
| True True Char
System.ValueType
|
| [C:\psh]
| 124> invoke-expression 3${a}4
| 7
| [C:\psh]
| 125> invoke-expression 3${a[0]}4
| 34
| [C:\psh]
|
|

> Trying to add 2 different numbers together using something like:
> invoke-expression 3${a}4
>
> What I'd really like to have working though is something like this:
> invoke-expression 3${a[0]}4
>
> Now, $a.gettype() and $a[0].gettype() come out differently, and I don't
> know if/how to make $a[0] into a string like $a would be.

Fixed...

I just added an extra bit of logic to workaround this:
$b=[string]$a[0]

Marco

Is this more along the lines of what you're looking for?

PS (1) > $ops = "+","-","*","/" # array of operators
PS (2) > 0..3 | foreach { invoke-expression "9 $($ops[$_]) 4" } # evaluate
the expression for each operator in the array...
13
5
36
2.25
PS (3) >

If you want to expand an expression in a string that is more complex than a
simple variable, then you have to put the expression in $( ... ).

Of course you could also do it this way :-)

PS (4) > foreach ($op in $ops) { invoke-expression "9 $op 4" }
13
5
36
2.25
PS (5) >

-bruce

--
Bruce Payette [MSFT]
Windows PowerShell Technical Lead
Microsoft Corporation
This posting is provided "AS IS" with no warranties, and confers no rights.

Visit the Windows PowerShell Team blog at:
http://blogs.msdn.com/PowerShell
Visit the Windows PowerShell ScriptCenter at:
http://www.microsoft.com/technet/scr.../hubs/msh.mspx
My Book: http://manning.com/powershell

"Marco Shaw" <marcoDOTshaw_@_gmailDOTcom> wrote in message
news:eIuuEiWUHHA.4276@TK2MSFTNGP02.phx.gbl...
>> Trying to add 2 different numbers together using something like:
>> invoke-expression 3${a}4
>>
>> What I'd really like to have working though is something like this:
>> invoke-expression 3${a[0]}4
>>
>> Now, $a.gettype() and $a[0].gettype() come out differently, and I don't
>> know if/how to make $a[0] into a string like $a would be.
>
> Fixed...
>
> I just added an extra bit of logic to workaround this:
> $b=[string]$a[0]
>
> Marco
>

Speaking of the range operator... It seems to me, I would like to use it
like so:
$a = 1,2,30..50, 60..100

Likewise:
$a[1,3..5,9..100]

--
William Stacey [C# MVP]


"Bruce Payette [MSFT]" <brucepay@microsoft.com> wrote in message
news:OBeegLZUHHA.4796@TK2MSFTNGP05.phx.gbl...
| Is this more along the lines of what you're looking for?
|
| PS (1) > $ops = "+","-","*","/" # array of operators
| PS (2) > 0..3 | foreach { invoke-expression "9 $($ops[$_]) 4" } # evaluate
| the expression for each operator in the array...
| 13
| 5
| 36
| 2.25
| PS (3) >
|
| If you want to expand an expression in a string that is more complex than
a
| simple variable, then you have to put the expression in $( ... ).
|
| Of course you could also do it this way :-)
|
| PS (4) > foreach ($op in $ops) { invoke-expression "9 $op 4" }
| 13
| 5
| 36
| 2.25
| PS (5) >
|
| -bruce
|
| --
| Bruce Payette [MSFT]
| Windows PowerShell Technical Lead
| Microsoft Corporation
| This posting is provided "AS IS" with no warranties, and confers no
rights.
|
| Visit the Windows PowerShell Team blog at:
| http://blogs.msdn.com/PowerShell
| Visit the Windows PowerShell ScriptCenter at:
| http://www.microsoft.com/technet/scr.../hubs/msh.mspx
| My Book: http://manning.com/powershell
|
| "Marco Shaw" <marcoDOTshaw_@_gmailDOTcom> wrote in message
| news:eIuuEiWUHHA.4276@TK2MSFTNGP02.phx.gbl...
| >> Trying to add 2 different numbers together using something like:
| >> invoke-expression 3${a}4
| >>
| >> What I'd really like to have working though is something like this:
| >> invoke-expression 3${a[0]}4
| >>
| >> Now, $a.gettype() and $a[0].gettype() come out differently, and I don't
| >> know if/how to make $a[0] into a string like $a would be.
| >
| > Fixed...
| >
| > I just added an extra bit of logic to workaround this:
| > $b=[string]$a[0]
| >
| > Marco
| >
|
|

"William Stacey [C# MVP]" <william.stacey@gmail.com> wrote in message
news:eUOldRZUHHA.1208@TK2MSFTNGP03.phx.gbl...
> Speaking of the range operator... It seems to me, I would like to use it
> like so:
> $a = 1,2,30..50, 60..100
>
> Likewise:
> $a[1,3..5,9..100]
>

You can. Try this:

1,2+30..50+60..100

--
Keith

I saw that, but that made me feel dirty :-) As an operator, not sure why
this could not be made to work as I shown. Good workaround, thanks.

--
William Stacey [C# MVP]


"Keith Hill" <r_keith_hill@mailhot.nospamIdotcom> wrote in message
news:0B7F9BD0-CCA2-4686-9D39-40B8E99E0432@microsoft.com...
| "William Stacey [C# MVP]" <william.stacey@gmail.com> wrote in message
| news:eUOldRZUHHA.1208@TK2MSFTNGP03.phx.gbl...
| > Speaking of the range operator... It seems to me, I would like to use
it
| > like so:
| > $a = 1,2,30..50, 60..100
| >
| > Likewise:
| > $a[1,3..5,9..100]
| >
|
| You can. Try this:
|
| 1,2+30..50+60..100
|
| --
| Keith
|

> If you want to expand an expression in a string that is more complex than
> a simple variable, then you have to put the expression in $( ... ).
>
> Of course you could also do it this way :-)
>
> PS (4) > foreach ($op in $ops) { invoke-expression "9 $op 4" }

Thanks Bruce. That's nice to know.

I really have to sit down and read your book!