System.FilePath.joinDrive problem

[Twan van Laarhoven]

Bulat Ziganshin wrote:

> Hello Twan,
> 
> Saturday, November 3, 2007, 8:56:33 PM, you wrote:
> 
> 
>>This seems wrong, as far as I know "C:File" is not a valid path. What is
>>the reason for this special case?
> 
> 
> it's correct path and i think that program has correct behavior in
> this case. if you need to construct "c:\file", you should use "c:\" as
> directory (written as "c:\\", of course)
> 

On windows the 'path' "C:File" refers to the file named "File" on the 
drive "C:", in the *current directory* for that drive. It is a kind of 
half relative path.

This behaviour is at the very least unintuitive. In my opinion a 'path' 
like "C:File" is completely useless. The FilePath library never produces 
"C:" as an element itself, so what we do here doesn't matter much. A 
good argument in favor of adding the backslash is that a trailing slash 
on a directory name should not matter. I.e, for f not ending in a slash:
    (f </> g) == (f ++ [pathSeparator] </> g)
This holds for normal directories, but not if f is a drive letter.

I have investigated what some different platforms do:
  - Win32: PathAppend           gives "C:\File"
  - .Net:  Path.Combine         gives "C:File"
  - Perl:  File::Spec->catfile  gives "C:/File"

I would take the behaviour of the Win32 api as authorative here.

Twan