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How to get argc/argv in CWinApp?

Posted on 1998-10-19
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Last Modified: 2013-11-20
I would like to be able to get command line options in
argc & argv format in my InitInstance() call without
having to parse through the m_lpCmdLine string myself.
Is there an easy way to do this?
Thanks.
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Question by:bwilhelm
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Use __argv and __argc (double underline precedes the name) :

(begin quote)
Windows-based and Win32-based application use WinMain() as the entry point. WinMain() has the lpszCmdLine parameter for the command line arguments, but this parameter does not include the program name. However, you can

 still get to the program name through the variable __argv.

In a 32-bit application you can just reference __argv directly. It is declared in stdlib.h
In a 16-bit application, __argv is not declared and you will need to declare it. Include the following declaration in your source or header file:

#ifdef __cplusplus extern "C"  #endif extern char ** __argv;

To test this out, you can view __argv[0] by including the following line in your application:


   MessageBox( NULL, __argv[0], "__argv[0]", MB_OK );

NOTE: argv[0] will contain a pointer to the program name as it was executed from the command line or Program Manager: either qualified or unqualified.
(end quote)

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