HOWTO: Obtain the Program Name in a Windows-Based Application

ID: Q126571

The information in this article applies to:

Microsoft Visual C++ for Windows, 16-bit edition, versions 1.0, 1.5, 1.51, 1.52
Microsoft Visual C++, 32-bit Editions, versions 1.0, 2.0, 2.1, 4.0, 5.0, 6.0

SUMMARY

MS-DOS applications written in C generally use main() for the entry point. The command line can be obtained from the argv parameter of main(). In particular, the program name is pointed to by argv[0].

There is a similar mechansim that can be used from Windows-based and Win32-based applications.

MORE INFORMATION

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.

REFERENCES

The Windows 3.1 SDK "Guide to Programming", section 14.3.

Additional query words:

Keywords : kbLangC kbVC kbVC100 kbVC150 kbVC151 kbVC152 kbVC200 kbVC210 kbVC400 kbVC500 kbVC600
Version : WIN3X:1.0,1.5,1.51,1.52;WINNT:1.0,2.0,2.1,4.0,5.0;
Platform : NT WINDOWS
Issue type : kbhowto