Why won't generic accept type param?

[Brad Wood]

The following works fine; passing a type to CreateInstance:
private Type _next;
BaseWizardForm nextForm = (BaseWizardForm)Activator.CreateInstance(
_next );

So why doesn't this work (error = _next' is a 'field' but is used like a
'type'):
BaseWizardForm nextForm =
(BaseWizardForm)Activator.CreateInstance<_next>();

_next *is* a type...

And this gives the error "type or namespace name 'corndog' could not be
found.":
Type corndog = typeof(WelcomeForm);
BaseWizardForm nextForm =
(BaseWizardForm)Activator.CreateInstance<corndog>( );

It gives that error on the second line...

[Mattias Sjögren]

>_next *is* a type...

Actually it's an instance of System.Type, something only known at
runtime. That's not what the compiler considers to be a type. So you
have to use Reflection to construct the generic type at runtime.
Mattias

--
Mattias Sjögren [C# MVP] mattias @ mvps.org
http://www.msjogren.net/dotnet/ | http://www.dotnetinterop.com
Please reply only to the newsgroup.

[Jon Skeet [C# MVP]]

Brad Wood <du**@sweet.comwrote:

The following works fine; passing a type to CreateInstance:
private Type _next;
BaseWizardForm nextForm = (BaseWizardForm)Activator.CreateInstance(
_next );

So why doesn't this work (error = _next' is a 'field' but is used like a
'type'):
BaseWizardForm nextForm =
(BaseWizardForm)Activator.CreateInstance<_next>();

_next *is* a type...

And this gives the error "type or namespace name 'corndog' could not be
found.":
Type corndog = typeof(WelcomeForm);
BaseWizardForm nextForm =
(BaseWizardForm)Activator.CreateInstance<corndog>( );

It gives that error on the second line...

You're trying to give information at runtime which the compiler needs
at compile time. It's like trying to use:

Type corndog = typeof(WelcomeForm);
corndog f = new corndog();

The compiler doesn't know what type you're talking about, so it can't
do the *compile-time* checks/references it needs to.

--
Jon Skeet - <sk***@pobox.com>
http://www.pobox.com/~skeet Blog: http://www.msmvps.com/jon.skeet
If replying to the group, please do not mail me too

[Michael C]

"Jon Skeet [C# MVP]" <sk***@pobox.comwrote in message
news:MP************************@msnews.microsoft.c om...

You're trying to give information at runtime which the compiler needs
at compile time. It's like trying to use:

Type corndog = typeof(WelcomeForm);
corndog f = new corndog();

The compiler doesn't know what type you're talking about, so it can't
do the *compile-time* checks/references it needs to.

This is my understanding of how it works (correct me if i'm wrong). If you
use a generic class many times but only with int and string then the
compiler will compile 2 seperate copies of the class, one as if the
appropriate fields/params etc had been defined as int and the other as if
they had been defined as string. Because of this it can't take the type at
runtime.

Michaek

[Jon Skeet [C# MVP]]

Michael C wrote:

You're trying to give information at runtime which the compiler needs
at compile time. It's like trying to use:

Type corndog = typeof(WelcomeForm);
corndog f = new corndog();

The compiler doesn't know what type you're talking about, so it can't
do the *compile-time* checks/references it needs to.

This is my understanding of how it works (correct me if i'm wrong). If you
use a generic class many times but only with int and string then the
compiler will compile 2 seperate copies of the class, one as if the
appropriate fields/params etc had been defined as int and the other as if
they had been defined as string. Because of this it can't take the type at
runtime.

No, the compiler doesn't create the copies of the class - the CLR does,
and indeed you *can* create generic classes using reflection at
runtime. The problem is that the compiler wants to know about the type
at compile-time - it just doesn't have enough information to work on in
the above scenario.

Jon

[Brad Wood]

"Mattias Sjögren" <ma********************@mvps.orgwrote in message
news:eV**************@TK2MSFTNGP04.phx.gbl...

_next *is* a type...

Actually it's an instance of System.Type, something only known at
runtime. That's not what the compiler considers to be a type. So you
have to use Reflection to construct the generic type at runtime.

Then I have no idea why Activator has a generic overload. Or maybe I'm just
completely confused. It seems you would always want to pass to Activator a
Type variable rather than explictly referring to the class name. If you
know the class name at compile time, you would just use it's constructor,
not Activator. Right?

[Mattias Sjögren]

Brad,

>Then I have no idea why Activator has a generic overload. Or maybe I'm just
completely confused. It seems you would always want to pass to Activator a
Type variable rather than explictly referring to the class name. If you
know the class name at compile time, you would just use it's constructor,
not Activator. Right?

One reason for the generic Activator.CreateInstance method is in fact
to implement the new operator. This code

class Foo<Twhere T : new() {
T t = new T();
}

compiles to

class Foo<Twhere T : new() {
T t = Activator.CreateInstance<T>();
}
Mattias

--
Mattias Sjögren [C# MVP] mattias @ mvps.org
http://www.msjogren.net/dotnet/ | http://www.dotnetinterop.com
Please reply only to the newsgroup.

[Brad Wood]

"Mattias Sjögren" <ma********************@mvps.orgwrote in message
news:eM**************@TK2MSFTNGP05.phx.gbl...

One reason for the generic Activator.CreateInstance method is in fact
to implement the new operator. This code

class Foo<Twhere T : new() {
T t = new T();
}

compiles to

class Foo<Twhere T : new() {
T t = Activator.CreateInstance<T>();
}

I think it took this example for the light bulb to go off in realizing the
difference between compile time substitiution of generic arguments and run
time evaluation of type variables...