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?
In DOS and Windows programming, 16 bits is a "WORD", 32 bits is a "DWORD" (double word), and 64 bits is a "QWORD"; but in other contexts "word" means the machine's natural binary processing size, which ranges from 32 to 64 bits nowadays.
A DWORD is a 32-bit unsigned integer (range: 0 through 4294967295 decimal).
Hope that helps. Show activity on this post. No ... on all Windows platforms DWORD is 32 bits. LONGLONG or LONG64 is used for 64 bit types.
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.
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