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?
问题:
回答1:
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).
回答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.
回答3:
No ... on all Windows platforms DWORD is 32 bits. LONGLONG or LONG64 is used for 64 bit types.
回答4:
8 bits is a byte. 2 bytes is a word. Double word or DWORD is 4 bytes or 2 words.
回答5:
DWord is a Double word and a word is 65535. Double Word would then be 65535^2 = 4294836225 or ulong and its 4 bytes in size word is 2 byte
And this is 2 bytes to store on harddrive or send over the internet.
回答6:
:) word on modern processors is either 32-bit or 64-bit. It's simply memory pointer's length (which is ALU's capacity in turn).
But historically x86 "word" is 16 bits (instead of 32). Thereby Microsoft libraries which historically target x86, define DWORD as unsigned long i.e. "machine pointer size".
That's all kids. For future reference see Wikipedia.
回答7:
Call a long, dword32, dword64... whatever you want, but a byte is 8 bits a word is 2 bytes a dword (double word) is 2 words a qword (quad word) is 4 words
On 32 or 64bit systems the aliases like LONG or INT may vary, but a dword is still a double word