包含联合的结构的大小 [英] Size of structs containing unions
问题描述
在Visula C ++ 6.0中我已经声明了这样的结构：
typedef struct _WRITE_INPUT {
ULONG DeviceNumber;
ULONG RegisterNumber;
union {
USHORT ShortData;
UCHAR CharData; < br $>
};
} WRITE_INPUT;
我不明白为什么，sizeof（WRITE_INPUT）返回12.它应该
返回10，不应该吗？
sizeof（ULONG）= 4
sizeof（ULONG）= 4
sizeof（USHORT）= 2（最长的工会领域）
4 + 4 + 2 = 10
谢谢
再见
luke
luke写道：hi，
我已经声明了这样的结构：
typedef struct _WRITE_INPUT {
ULONG DeviceNumber;
ULONG RegisterNumber ;
union {
USHORT ShortData;
UCHAR CharData;
};
} WRITE_INPUT;
我无法理解为什么，sizeof（WRITE_INPUT）返回12.它应该
返回10，不应该'是吗？
sizeof（ULONG）= 4
sizeof（ULONG）= 4
sizeof（USHORT）= 2（最长的工会领域）
> 4 + 4 + 2 = 10
谢谢
再见
luke
clc FAQ＃2.13会回答你的问题
谢谢，
我认为这是工会问题
再见
"路" < LR ***** @ yahoo.com>写道：Visula C ++ 6.0中的我已经声明了这样的结构：
typedef struct _WRITE_INPUT {
不要使用以下划线开头的标识符;他们保留了实施的
。 （它稍微复杂一些，但它只是为了避免它们而是最安全的。）
ULONG DeviceNumber;
ULONG RegisterNumber; <联盟{
USHORT ShortData;
UCHAR CharData;
};
} WRITE_INPUT;
我无法理解为什么，sizeof（ WRITE_INPUT）返回12.它应该返回10，不应该吗？
sizeof（ULONG）= 4
sizeof（ULONG）= 4
sizeof（ USHORT）= 2（最长的联合字段）
4 + 4 + 2 = 10
编译器可以在任何成员之后添加填充结构。
在这种情况下，它可能在末尾添加2个字节的填充
使结构的大小为4的倍数，所以如果您有一系列结构，ULONG成员
将正确对齐。
顺便提一下，名称ULONG，USHORT和UCHAR并不是特别的
很有帮助。我认为它们分别是unsigned long，
unsigned short和unsigned char的typedef（或宏？）。为什么不直接使用
名字？
-
Keith Thompson（The_Other_Keith） ks *** @ mib.org < http://www.ghoti.net/~kst>
圣地亚哥超级计算机中心< * GT; < http://users.sdsc.edu/~kst>
我们必须做点什么。这是事情。因此，我们必须这样做。
hi,
in Visula C++ 6.0 I have declared a struct like this:
typedef struct _WRITE_INPUT {
ULONG DeviceNumber;
ULONG RegisterNumber;
union {
USHORT ShortData;
UCHAR CharData;
};
} WRITE_INPUT;
I can''t understand why, sizeof(WRITE_INPUT) returns 12. It should
return 10, shouldn''t it?
sizeof(ULONG) = 4
sizeof(ULONG) = 4
sizeof(USHORT) = 2 (longest union field)
4 + 4 + 2 = 10
thanks
bye
luke
luke wrote:hi,
in Visula C++ 6.0 I have declared a struct like this:
typedef struct _WRITE_INPUT {
ULONG DeviceNumber;
ULONG RegisterNumber;
union {
USHORT ShortData;
UCHAR CharData;
};
} WRITE_INPUT;
I can''t understand why, sizeof(WRITE_INPUT) returns 12. It should
return 10, shouldn''t it?
sizeof(ULONG) = 4
sizeof(ULONG) = 4
sizeof(USHORT) = 2 (longest union field)
4 + 4 + 2 = 10
thanks
bye
luke
c.l.c FAQ #2.13 would answer your question
thank you,
I thought it was a union problem
bye
"luke" <lr*****@yahoo.com> writes:in Visula C++ 6.0 I have declared a struct like this:
typedef struct _WRITE_INPUT {
Don''t use identifiers starting with an underscore; they''re reserved to
the implementation. (It''s slightly more complex than that, but it''s
safest just to avoid them.)
ULONG DeviceNumber;
ULONG RegisterNumber;
union {
USHORT ShortData;
UCHAR CharData;
};
} WRITE_INPUT;
I can''t understand why, sizeof(WRITE_INPUT) returns 12. It should
return 10, shouldn''t it?
sizeof(ULONG) = 4
sizeof(ULONG) = 4
sizeof(USHORT) = 2 (longest union field)
4 + 4 + 2 = 10
The compiler is free to add padding after any member of a structure.
In this case, it''s probably adding 2 bytes of padding at the end to
make the size of the structure a multiple of 4, so the ULONG members
will be aligned properly if you have an array of structures.
Incidentally, the names ULONG, USHORT, and UCHAR aren''t particularly
helpful. I presume they''re typedefs (or macros?) for unsigned long,
unsigned short, and unsigned char, respectively. Why not just use the
names directly?
--
Keith Thompson (The_Other_Keith) ks***@mib.org <http://www.ghoti.net/~kst>
San Diego Supercomputer Center <*> <http://users.sdsc.edu/~kst>
We must do something. This is something. Therefore, we must do this.
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