Python openfile returing IOError22, even with escaped filename

Question

1. I tried to create a new file with Python26 file()/open() directly inside C:\. User does not have permission to create a new file here.

2. Expectation is that i receive a "Permission denied" error. However i end up receiving invalid file name/mode error on one particular host. This is breaking the error handling of my application . On all other hosts it is doing fine.

3. All other threads in stack over flow like Region: IOError: [Errno 22] invalid mode ('w') or filename point to wrong file names which i believe is not the case with me.

   a = file("C:\\new2.txt","w")                                  
   Traceback (most recent call last):  
   File "<stdin>", line 1, in <module>                             
   IOError: [Errno 22] invalid mode ('w') or filename: 'C:\\new2.txt'
   

Answer 1

Whey you are calling file(path,"w") in windows, Python calls open() or _wopen() with flags O_CREAT | O_TRUNC.

Both this functions can return following codes (as described in MSDN):

• EACCES - Tried to open a read-only file for writing, file's sharing mode does not allow the specified operations, or the given path is a directory.
• EEXIST - _O_CREAT and _O_EXCL flags specified, but filename already exists.
• EINVAL - Invalid oflag or pmode argument.
• EMFILE - No more file descriptors are available (too many files are open).
• ENOENT - File or path not found.

As you can see, EACCES is not your option as file does not exists.

Now let's take a look at flags:

• _O_TRUNC Opens a file and truncates it to zero length; the file must have write permission. Cannot be specified with _O_RDONLY. _O_TRUNC used with _O_CREAT opens an existing file or creates a file.

And one last thing:

if any value other than the allowed oflag values is specified, the function generates an assertion in Debug mode and invokes the invalid parameter handler, as described in Parameter Validation. If execution is allowed to continue, the function returns -1 and sets errno to EINVAL.

So, in your case you will not get EACESS. Instead you will get EINVAL (22).

You can check this question How to check if a file can be created inside given directory on MS XP/Vista? and this Python - Test directory permissions

To try to find solution. As you can see, Windows file access rights hardly can be mapped to POSIX. There are no easy solution.