CODEWITHSUNDEEP
visual-studiowinapivisual-c++shellexecute
vico
vico

Reputation: 18171

Start external program in it's directory

I need to start the external program c:\pro\prog1.exe from my project. The external program has its configuration file in the same directory c:\pro\prog1.ini. I do:

ShellExecute(NULL,L"open",L"c:\pro\prog1.exe" ,NULL,NULL,SW_SHOWDEFAULT);

Program c:\pro\prog1.exe starts, but it does not load its configuration file c:\pro\prog1.ini. It looks like I need to place the .ini file in the same directory where my host application is run from. This is not acceptable. So, how to start an external program and ask Windows to run it from its directory?

Upvotes: 1

Views: 828

Answers (1)

Barmak Shemirani
Barmak Shemirani

Reputation: 31599

The 5th parameter in ShellExecute is the startup directory.

Alternatively "prog.exe" can use GetModuleFileName and PathRemoveFileSpec to find its own directory, as suggested in comments.

Note that some directories like "c:\\Program Files" and "c:\\Program Files (x86)" require elevated access to create/modify/delete files (for example during installation). A process without elevated access, can access files in protected directories with read-only flag. Otherwise Windows will redirect the path to a different directory if write-access is requested.

For normal execution, the *.exe should use "Documents" or "AppData" folder to read/write data.

Upvotes: 2

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