Re: where __declspec(dllexport) is defined?

From:
"Ben Voigt [C++ MVP]" <rbv@nospam.nospam>
Newsgroups:
microsoft.public.vc.language
Date:
Fri, 26 Dec 2008 17:53:46 -0600
Message-ID:
<221B286F-EC45-472D-B998-71B5FEE7D624@microsoft.com>
"GiuseppeDini" <vvene@libero.it> wrote in message
news:0fe3b34c-c963-49a4-b480-d8267f889185@t26g2000prh.googlegroups.com...

On 9 Dic, 22:11, "Doug Harrison [MVP]" <d...@mvps.org> wrote:

On Tue, 9 Dec 2008 13:04:10 -0800 (PST), GiuseppeDini <vv...@libero.it>
wrote:

I created a dll project named "test" and I've found this code:

#ifdef TEST_EXPORTS
#define TEST_API __declspec(dllexport)
#else
#define TEST_API __declspec(dllimport)
#endif

There is one thing i can't understand.
Where is TEST_EXPORTS defined when the code is compiled in the dll to
be exported?
I can't find a line where it is defined.


Look in your preprocessor options (/D).


Thank you.
I understand that it is a basic knowledge, but I'm new to c++.

I've noticed several names for that option:
WIN32;_DEBUG;_WINDOWS;_USRDLL;CCCC_EXPORTS.
So my application is also defined for win32. If I select a win64
project, the same code will run under a 64bit processor and one of the
above options will become WIN64?
(but I have vc++ express and I can't try it)


I think 64-bit builds define *both* WIN32 and WIN64 (and maybe also _WIN32
and _WIN64)

   Giuseppe

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