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c++winapi64-bitdword
Haim Bender
Haim Bender

Reputation: 8187

How large is a DWORD with 32- and 64-bit code?

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?

Upvotes: 62

Views: 163290

Answers (4)

ogggre
ogggre

Reputation: 2266

Windows API defines DWORD sizes as follows:

  • x86: sizeof(DWORD) = 4
  • x64: sizeof(DWORD) = 4

Upvotes: 1

Nir
Nir

Reputation: 29594

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).

Upvotes: 72

Mark Ingram
Mark Ingram

Reputation: 73625

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.

Upvotes: 18

Rob Walker
Rob Walker

Reputation: 47462

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

Upvotes: 13

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