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497

answers:

4

I'm not afraid to admit that I'm somewhat of a C++ newbie, so this might seem like a silly question but....

I see DWORD used all over the place in code examples. When I look up what a DWORD truly means, its apparently just an unsigned int (0 to 4,294,967,295). So my question then is, why do we have DWORD? What does it give us that the integral type 'unsigned int' does not? Does it have something to do with portability and machine differences?

+17  A: 

DWORD is not a C++ type, it's defined in <windows.h>.

The reason is that DWORD has a specific range and format Windows functions rely on, so if you require that specific range use that type. (Or as they say "When in Rome, do as the Romans do.") For you, that happens to correspond to unsigned int, but that might not always be the case. To be safe, use DWORD when a DWORD is expected, regardless of what it may actually be.

For example, if they ever changed the range or format of unsigned int they could use a different type to underly DWORD to keep the same requirements, and all code using DWORD would be none-the-wiser. (Likewise, they could decide DWORD needs to be unsigned long long, change it, and all code using DWORD would be none-the-wiser.)


Also note unsigned int does not necessary have the range 0 to 4,294,967,295. See here.

GMan
+1, that DWORD may change sometime in the future.
Default
Yea I knew it wasn't a type, and I figured it had something to do with portability and decoupling it from its underlying data type representation, but you've filled in a number of holes for me. Thanks!
byte
+1 the explanation. But I still would bet that if they did change DWORD a lot of programs would still break :)
Akusete
@Akusete: I agree. I doubt they ever actually would, especially at such a late stage. :)
GMan
+2  A: 

SDK developers prefer to define their own types using typedef. This allows to change underlying types only in one place, without changing all client code. It is important to follow this convention. DWORD is unlikely to be changed, but types like DWORD_PTR are different in different platforms, like Win32 and x64. So, if some function has DWORD parameter, use DWORD and not unsigned int, and your code will be compiled in all future windows headers versions.

Alex Farber
A: 

For myself, I would assume unsigned int is platform specific. Integer could be 8 bits, 16 bits, 32 bits or even 64 bits.

DWORD in the other hand, specifies its own size, which is Double Word. Word are 16 bits so DWORD will be known as 32 bit across all platform

YeenFei
8-bit `int` isn't allowed under the C or C++ standards. But the point remains that there are platforms with 16-bit `int` and others with 32-bit `int`.
dan04
"Word are 16 bits" made sense in 1984 but really doesn't anymore :-(
Zack
+8  A: 

When MS-DOS and Windows 3.1 operated in 16-bit mode, an Intel 8086 word was 16 bits, a Microsoft WORD was 16 bits, a Microsoft DWORD was 32 bits, and a typical compiler's unsigned int was 16 bits.

When Windows NT operated in 32-bit mode, an Intel 80386 word was 32 bits, a Microsoft WORD was 16 bits, a Microsoft DWORD was 32 bits, and a typical compiler's unsigned int was 32 bits. The names WORD and DWORD were no longer self-descriptive but they preserved the functionality of Microsoft programs.

When Windows operates in 64-bit mode, an Intel word is 64 bits, a Microsoft WORD is 16 bits, a Microsoft DWORD is 32 bits, and a typical compiler's unsigned int is 32 bits. The names WORD and DWORD are no longer self-descriptive, AND an unsigned int no longer conforms to the principle of least surprises, but they preserve the functionality of lots of programs.

I don't think WORD or DWORD will ever change.

Windows programmer
Good history information. +1
byte