In Visual C++ a DWORD is just an unsigned long that is machine, platform, and SDK dependent. However, since DWORD is a double word (that is 2 * 16), is a DWORD still 32-bit on 64-bit architectures?
Actually, on 32-bit computers a word is 32-bit, but the DWORD type is a leftover from the good old days of 16-bit.
In order to make it easier to port programs to the newer system, Microsoft has decided all the old types will not change size.
You can find the official list here: http://msdn.microsoft.com/en-us/library/aa383751(VS.85).aspx
All the platform-dependent types that changed with the transition from 32-bit to 64-bit end with _PTR (DWORD_PTR will be 32-bit on 32-bit Windows and 64-bit on 64-bit Windows).
It is defined as:
typedef unsigned long DWORD;
However, according to the MSDN:
On 32-bit platforms, long is synonymous with int.
Therefore, DWORD is 32bit on a 32bit operating system. There is a separate define for a 64bit DWORD:
typdef unsigned _int64 DWORD64;
Hope that helps.
No ... on all Windows platforms DWORD is 32 bits. LONGLONG or LONG64 is used for 64 bit types.
来源：https://stackoverflow.com/questions/39419/how-large-is-a-dword-with-32-and-64-bit-code