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I was trying to compile and project file ,taken freshly from the VSS and it was throwing an error

          could not create output file" release\emsDataAccess.dcu " .

Strangely the error was pointing to line "90" where as the last line of code in the error-file (emsDataAccess.pas) is"75".

Kindly help! Regards -Vas

2 Answers 2

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Does your release subdirectory exist? The compiler won't create it for you, and if it's not in the VSS repository chances are it's missing in a fresh checkout.

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3 Comments

yes there is an empty release sub-directory in the VSS named "release". So I have to created a sub-directory in the folder "release" in my current project folder, and it worked. Thanks a bunch mghie. Have a nice day!
i was observing in my VSS , that not all the Projects have an "release subdirectory "...only some projects have them . whats the criteria/guidelines when a project needs an release subdirectory? Do have to interact with the DB?
vas: You should consult with the creator / owner of the projects in VSS. There are no rules, just best practices. However, all directories referenced in the project need to be there, either directly from the checkout, or from a helper script preparing the build after checkout.
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It probably points you to "line 90" of the build script.

It's simply a "cannot create file" error, so either the Path is invalid, you don't have the rights to create files or folders or maybe emsDataAccess.dcu was checked out as read-only.

Just use explorer to see if the folder and/or file already exist. Delete the file if you find it.

3 Comments

And we should note that you should never check DCU files into source control.
it is also suggested that .pas files should always be saved in VSS and DCU files never check them to VSS..... any reason for that?
Because the PAS files are your actual source code. That's what source control is about. Checking binaries in, such as DCU files, isn't necessary if they can be produced from source files. Also, it's a heck of a lot easier to run a diff on a text file than on a binary.

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