Typically, what does it mean when java returns a "The parameter is incorrect"

Question

I am creating a file like so

try {
    File file = new File(workingDir, obj.getName() + ".xls");
    outputStream = new FileOutputStream(file);
} catch (FileNotFoundException ex) {
    ex.printStackTrace();
}

And I am getting

java.io.FileNotFoundException: ..\a\relative\path\obj_name.xls (The parameter is incorrect)
    at java.io.FileOutputStream.open(Native Method)
    at java.io.FileOutputStream.<init>(Unknown Source)
    at java.io.FileOutputStream.<init>(Unknown Source)

What does "the parameter is incorrect" mean? The pathname I validated exists; shouldn't it just create the file name? This is on a windows machine. The code works without this error on unix based systems.

---

Update

Verified that the file exists that the output stream is attempting to write to. Also verified that the file is writable. After much fussing, I removed the actual path and just passed in the file name (not desired) and that works. So the issue has something to do with the path. Do I need to escape the characters in the path?

Answer 1

Perhaps the application doesn't have the correct access to write to the file? Is it read-only or otherwise protected?

FileOutputStream.open() is a native method, I would assume any sort of exception message such as "The parameter is incorrect" is coming from the underlying OS.

BTW: the File constructor does not call FileOutputStream.open(), so is the exception not actually coming from the code you posted here?

Answer 2

Maybe it's because of the backslashes in the path? Path too long? File name invalid for this error (special caracters...) ?

I might be totally wrong, but worth a try since it sounds like a OS dependent error.

Answer 3

This looks like a reported bug on Windows machines.

Under normal situations, something like a path that has a colon (:) in it which does not refer to a drive letter would cause this message.

Answer 4

It appears to be an issue with the path you're using. Try using file.getPath() before you open it to debug what is going on with your path.

File file = new File(workingDir, obj.getName() + ".xls");
System.out.println("Path="+file.getPath());
outputStream = new FileOutputStream(file);

Answer 5

If your "workingDir" is a relative path, then are you sure you are on the correct "current directory" when you moved your app from unix to windows? Maybe, you should check what the current directory of the running application is.

Answer 6

Make sure the user that runs the JVM process has the right permissions to access that file.