#usind <mscorlib.dll> does not work ?

I

iceColdFire

Hi @all.

I am working on a c++ module. I am trying to compile one of the
programs containing a statement of type

#using <mscorlib.dll>

however, there seems to be some problem...#using directive illegal is
the error.
Compiler I am using is MS VC++ version 6.0
Platform is Intel 32 / Windows 2000

Suggestions are requested on the problem...

Thanks,
a.a.cpp
 
P

__PPS__

iceColdFire said:
Hi @all.

I am working on a c++ module. I am trying to compile one of the
programs containing a statement of type

#using <mscorlib.dll>

it has nothing to do with c++, c# maybe?
 
I

iceColdFire

::__PPS__

perfect...ok...can you explain what exactly it means in C#. and does
C++ MSVC Version 6.0 include or provide any other operator for similar
purpose.

Thanks,
/\/\cpp
 
R

Rade

I am working on a c++ module. I am trying to compile one of the
programs containing a statement of type

#using <mscorlib.dll>

This is out of topic, but FYI the #using directive is used to import
metadata from .NET assemblies to a C++ program, so it requires at least
Visual C++ 7.0 or later.

Rade
 
W

Will

however, there seems to be some problem...#using directive illegal is
the error.
Compiler I am using is MS VC++ version 6.0
Platform is Intel 32 / Windows 2000

using is a C# directive for including some of the managed namespaces, like
using System.Security; in managed C++ you can #include <mscorlib.dll> but
that's the closest I've seen to what your trying to do, this really should
be directed to a M$ group as its not strictly C++.
 
R

Richard Herring

Rade said:
This is out of topic, but FYI the #using directive is used to import
metadata from .NET assemblies to a C++ program, so it requires at least
Visual C++ 7.0 or later.
Equally off-topic, but it's also used for importing metadata from COM
type libraries, which works in VC++ 6.
 
R

Rade

Equally off-topic, but it's also used for importing metadata from COM type
libraries, which works in VC++ 6.

That is #import, not #using...

(I am replying just for the sake of truth, but I'd like to close the
M$-specific discussion now.)

Rade
 

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