Visual C++: How large is a DWORD with 32- and 64-bit code?

Question

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?

Answer 1

No ... on all Windows platforms DWORD is 32 bits. LONGLONG or LONG64 is used for 64 bit types.

Answer 2

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.

Answer 3

:) word on moden processors is ether 32 or 64. It's simply memory pointer's length (witch is ALU's capacity in turn).

But historically x86 "word" is 16 bits (insted of 32). Thereby Microsoft libraries witch was historically targeting on x86 define DWORD as unsigned long i.e. "machine pointer size".

That's all kids. For future reference see Wikipedia.

Answer 4

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).