CreateFile() ERROR_SHARING_VIOLATION with single thread - c++

Question

Why can't a single thread open the same file tqize using CreateFile and an exclusive file lock to the process? The example below will fail on the second attempt to open the file by the same thread with a ERROR_SHARING_VIOLATION exception:
ERROR_SHARING_VIOLATION 32 (0x20) The process cannot access the file
because it is being used by another process.
Emphasis on the word "process" above; it is the same process (and even the same thread) that tries to open the same file twize.
#include <Windows.h>
int _tmain(int argc, _TCHAR* argv[])
{
HANDLE hOutputFile1 = CreateFile(
// File name
L"test.dat",
// Requested access to the file
GENERIC_READ | GENERIC_WRITE,
// Share mode. 0 equals exclusive lock for the process
0,
// Pointer to a security attribute structure
NULL,
// Action to take on file
OPEN_ALWAYS,
// File attributes and flags
FILE_ATTRIBUTE_NORMAL,
// Template file
NULL
);
if (hOutputFile1 == INVALID_HANDLE_VALUE)
{
// Error
DWORD lastError = GetLastError();
return (int)lastError;
}
// opening the same file for the second time will return a ERROR_SHARING_VIOLATION exception
HANDLE hOutputFile2 = CreateFile(
// File name
L"test.dat",
// Requested access to the file
GENERIC_READ | GENERIC_WRITE,
// Share mode. 0 equals exclusive lock by the process
0,
// Pointer to a security attribute structure
NULL,
// Action to take on file
OPEN_ALWAYS,
// File attributes and flags
FILE_ATTRIBUTE_NORMAL,
// Template file
NULL
);
if (hOutputFile2 == INVALID_HANDLE_VALUE)
{
// Error
DWORD lastError = GetLastError();
return (int)lastError;
}
return 0;
}

Answer 1

The error message text is a bit misleading, but the fact that the two calls to CreateFile are made from the same thread in the same process doesn't change anything. Once the first call to CreateFile has been made, then subsequent calls to CreateFile, irrespective of where they originate, must obey the sharing rules.
I guess the error message text tries to capture the most common source of a sharing violation. Namely two processes competing for the same file. But the simple fact is that once you have opened a file with exclusive sharing, then no other attempts to open the file can succeed.

Answer 2

That old message is misleading. It doesn't matter what process or thread is opening a file. File sharing is handle-based.

Answer 3

From MSDN:
dwShareMode [in]:
If this parameter is zero and CreateFile succeeds, the file or device
cannot be shared and cannot be opened again until the handle to the
file or device is closed.
This is just talking about opening the file or device and the error message is something general and confusing.
Instead of trying to re-open the file, you should hold the handle and do your job.