I have some code that uses ctypes to try to determine if the file pointed to by sys.stdout is actually stdout. I know that on any POSIX-compliant system, and even on Windows, it should be safe to assume this is true if sys.stdout.fileno() == 1, so my question is not how to do this in general.
In my code (which is already using ctypes for something unrelated to my question) I carelessly had something like:
libc = ctypes.CDLL(ctypes.util.find_library('c'))
real_stdout = libc.fileno(ctypes.c_void_p.in_dll(libc, 'stdout'))
if sys.stdout.fileno() == real_stdout:
...
This works perfectly fine on Linux, so I didn't really think about it much. It looked nicer and more readable than hard-coding 1 as the file descriptor. But I found a few days later that my code wasn't working on OSX.
It turns outs OSX's libc doesn't export any symbol called 'stdout'. Instead its stdio.h has stdout defined as:
#define stdout __stdoutp
If I change my code to c_void_p.in_dll(libc, '__stdoutp') my code works as expected, but of course that's OSX-only. Windows, it turns out, has a similar issue (at least if using MSVC).
I will probably just change my code to use 1, but my question still stands, out of curiosity, if there's a cross-platform way to get the stdio pointer (and likewise stdin and stderr) without assuming that it's using the POSIX-compliant descriptor?
As so often when it comes to C, if you want compatibility, you'll have to go and look in the relevant standard. Since you mention windows, I guess you're not actually wanting the POSIX standard, but rather the C one.
C99 section 7,19,1 defines stdout to be a macro, and thus not a variable. That means there's no way you can rely on finding it using dlsym (which I assume in_dll uses). The actual expression could just as well be a function call or a fixed address. Perhaps not very likely, but it is possible...
As said in the comments, the fileno function is in turn defined by POSIX, not by C. C has no concept of file descriptors. I think you're better off assuming POSIX and just checking for the value 1, which it specifies.