Question

I have a program on a Windows XP computer on another continent. If I start it by double-clicking on an associated file, so that it runs the program according to the file-type association, it crashes at a certain point. If I start it by dragging an associated file to the program icon, it works fine. If I double-click on the icon and drag the associated file to the window, it works fine.

The guy on site assures me that he’s triple-checked the file-type association, and it should be correct.

Assuming that the icon points to the same executable as the registry, what differences would there be in starting the program in these two different ways?

EDIT: In response to a comment, the machine I’m having problems on is running in Japanese, while my normal machine is US English.

Answer 1

One possible difference is the initial working directory: if you drag a file onto the executable, the initial working directory is set to the directory containing the executable, whereas if you double click the file, the initial working directory is set to some default value independent of where the file or executable is located.

If you want to get consistent behavior in all cases, you can use SetCurrentDirectory() to set the current working directory to whatever you want; I recommend the directory containing the executable, which can be found by calling GetModuleFileName(NULL, ...) and stripping off the executable name, or by examining argv[0] inside main().