VC6：1 + 4 = 8为什么??? [英] VC6: 1+4=8 WHY???

问题描述

我有一个结构：

typedef struct {

signed char var1;

unsigned long int var2;

} MYSTRUCT;


现在看：

sizeof（签名字符）= 1（字节）

sizeof（unsigned long int）= 4（字节）


告诉我原因：

sizeof（MYSTRUCT）= 8 ??? （字节）

为什么它不是1 + 4 = 5个字节？


问候

-

Marcin P

GG1020924

skype：marcin_pil

I have a structure:
typedef struct {
signed char var1;
unsigned long int var2;
} MYSTRUCT;

Now look:
sizeof(signed char)=1 (byte)
sizeof(unsigned long int)=4 (bytes)

Tell me why:
sizeof(MYSTRUCT)=8 ??? (bytes)
Why it''s not 1+4=5 bytes?

Regards
--
Marcin P
GG1020924
skype: marcin_pil

推荐答案

Marcin P写道：

Marcin P wrote:

我有一个结构：
typedef struct {
signed char var1;
unsigned long int var2;
} MYSTRUCT;

告诉我原因：
sizeof（MYSTRUCT）= 8 ??? （字节）
为什么它不是1 + 4 = 5个字节？

I have a structure:
typedef struct {
signed char var1;
unsigned long int var2;
} MYSTRUCT;
Tell me why:
sizeof(MYSTRUCT)=8 ??? (bytes)
Why it''s not 1+4=5 bytes?

这是因为C ++编译器在var1之后添加了一些填充到

确保整个结构正确对齐。请参阅您的

编译器文档，并查看以下页面：

http://msdn.microsoft.com/library/de...m/msmod_18.asp

CrayzeeWulf

It is due to the fact that the C++ compiler adds some padding after var1 to
make sure that the entire structure is aligned correctly. Refer to your
compiler documentation and also check out the following page:

http://msdn.microsoft.com/library/de...m/msmod_18.asp

--
CrayzeeWulf

Marcin P< ma **** ********@poczta.onet.pl>在新闻中写道：4266db2b

Marcin P <ma************@poczta.onet.pl> wrote in news:4266db2b

1

@ news.vogel.pl：

1
@news.vogel.pl:

我有一个结构：
typedef struct {
signed char var1;
unsigned long int var2;
} MYSTRUCT;

现在看：
sizeof（signed char） = 1（字节）
sizeof（unsigned long int）= 4（字节）

告诉我原因：
sizeof（MYSTRUCT）= 8 ??? （字节）
为什么它不是1 + 4 = 5个字节？

I have a structure:
typedef struct {
signed char var1;
unsigned long int var2;
} MYSTRUCT;

Now look:
sizeof(signed char)=1 (byte)
sizeof(unsigned long int)=4 (bytes)

Tell me why:
sizeof(MYSTRUCT)=8 ??? (bytes)
Why it''s not 1+4=5 bytes?

要查找的几个概念：字节打包或结构对齐。


在你所使用的平台上，CPU可以更轻松地处理

字边界上的项目。因此，在您的结构中，它会尝试将您的成员放在单词

边界上。因此，你将有1个字节的var1,3个字节的填充，然后4个

个字节的var2，总大小为8个字节。

A couple of concepts to look up: byte packing, or structure alignment.

On whatever platform you''re on, the CPU can more easily address items on a
word boundary. So in your struct, it will try to put your members on word
boundaries. So, you''ll have 1 byte of var1, 3 bytes of filler, then 4
bytes of var2 for a total size of 8 bytes.

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