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Why in C++ do we use DWORD rather than unsigned int? [duplicate]

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?

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dreadwail Avatar asked Jun 08 '10 06:06

dreadwail


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Is dword signed or unsigned?

A DWORD is a 32-bit unsigned integer (range: 0 through 4294967295 decimal). Because a DWORD is unsigned, its first bit (Most Significant Bit (MSB)) is not reserved for signing.

What is the purpose of dword?

A dword, which is short for "double word," is a data type definition that is specific to Microsoft Windows. When defined in the file windows. h, a dword is an unsigned, 32-bit unit of data. It can contain an integer value in the range 0 through 4,294,967,295.

What dword means?

dword (plural dwords) (computing) A numerical value of twice the magnitude of a word, thus typically 32 bits.


4 Answers

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.

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GManNickG Avatar answered Oct 02 '22 08:10

GManNickG


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.

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Windows programmer Avatar answered Oct 01 '22 08:10

Windows programmer


SDK developers prefer to define their own types using typedef. This allows changing 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 on 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.

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Alex F Avatar answered Oct 01 '22 08:10

Alex F


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

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

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YeenFei Avatar answered Oct 02 '22 08:10

YeenFei