Dec 17, 2008
Every time I create a shortcut, usually by right-clicking on an executable and then choosing Create Shortcut from the context menu, the default path is always set to, for example, C:Program Files (x86)My Applicationmyapplication.exe. However, I already set my default Program Files and Program Files (x86) location to be in the D: drive. Nature of the Problem After installing Vista Ultimate x64, I installed TuneUp Utilities 2009 and from there, I set the default Program Files location to my second partition which is D:. I expected TuneUp Utilities to make the necessary changes to the registry. To make sure that no programs were left out in C:Program Files and C:Program Files (x86), I manually re-created the 2 folders in D: and then copied all the files and folders inside. I rebooted afterwards. Upon boot, my expectations toward TuneUp Utilities 2009 seemed to be correct because the default now points to D: (whenever I install 64-bit or 32-bit programs). I had no problems running the applications as well.
View 9 Replies
But a problem arises whenever a shortcut is made. It doesn't matter whether it was created by the installer itself or created manually. The target path always points to C: while the Start In path points to the correct one, which is D:. Whether it be in the Start Menu, on the Desktop, or in the Quick Launch menu, the same behavior happens. It's often a pain to edit the shortcuts manually- doing a right-click > Properties > change C to D > Change Icon > Apply > OK every time I install a new application or decide to create a shortcut for a certain portable app. I believe there's a registry entry that I can change so that the default Shortcut path now defaults to D: instead of C:.