使用指针数学访问结构元素 - 填充担心？ [英] Accessing structure elements using pointer math - padding worries?
问题描述
我可以在
的结构成员中使用指针算法吗？我应该担心结构填充吗？这在我的调试器中有效，但我想知道我是否在这里弯曲一些规则。
struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
struct char_struct test = {50,75,100,225};
unsigned char * c;
c =& test;
如果（c [2] == 100）//这没关系？
printf（" c [2] is 100 QUOT）; //这确实有效
c ++; //这没关系？
printf（" * c is 75"）; //这确实有效
Can I use pointer arithmetic on the members of a structure in the
following way? Should I be worried about structure padding? This works
in my debugger but I wonder if I''m bending some rule here.
struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
struct char_struct test = {50,75,100,225};
unsigned char *c;
c = &test;
if (c[2]==100) // this is okay?
printf("c[2] is 100"); // this does work
c++; // this is okay?
printf("*c is 75"); // this does work
推荐答案
qu ******** @ gmail.com 写道：
我可以对成员使用指针算法
中的结构如下？
Can I use pointer arithmetic on the members of a structure in the
following way?
是的，但不完全是你的意思。
Yes, but not quite as you mean.
我应该担心结构填充吗？
Should I be worried about structure padding?
是的。
Yes.
这在我的调试器中工作了
但是我想知道如果我在这里弯曲一些规则。
struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
struct char_struct test = { 50,75,100,225};
unsigned char * c;
c =& test;
如果（c [2] == 100）//这没关系？
This works
in my debugger but I wonder if I''m bending some rule here.
struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
struct char_struct test = {50,75,100,225};
unsigned char *c;
c = &test;
if (c[2]==100) // this is okay?
从你可以通过无符号
字符指针来解决任何问题，这是正常的，索引是少于
结构的大小。虽然它不一定能拿起结构的成员c
。
It''s OK in the sense that "you can address anything through an unsigned
character pointer", and the index is less that the size of the
structure. It doesn''t necessarily pick up member c of the structure
though.
printf（" c [2] is 100" ）; //这确实有效
printf("c[2] is 100"); // this does work
巧合，取决于编译器/平台。
By coincidence, depending on the compiler/platform.
c ++; //这没关系？
c++; // this is okay?
是的，只要你留在结构或步骤之内就可以超越
吧。
Yes, as long as you stay within the structure or step to just beyond
it.
printf（" * c is 75"）; //这确实有用
printf("*c is 75"); // this does work
再次，不保证是会员d。
Again, not guaranteed to be member d.
可以我在
的结构成员中使用指针算法？我应该担心结构填充吗？这在我的调试器中有效，但我想知道我是否在这里弯曲一些规则。
struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
struct char_struct test = {50,75,100,225};
unsigned char * c;
c =& test;
如果（c [2] == 100）//这没关系？
printf（" c [2] is 100 QUOT）; //这确实有效
c ++; //这没关系？
printf（" * c is 75"）; //这确实有效
following way? Should I be worried about structure padding? This works
in my debugger but I wonder if I''m bending some rule here.
struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
struct char_struct test = {50,75,100,225};
unsigned char *c;
c = &test;
if (c[2]==100) // this is okay?
printf("c[2] is 100"); // this does work
c++; // this is okay?
printf("*c is 75"); // this does work
不推荐这种访问方式。在编译器中没有添加任何
pad，因此您的测试程序可以正常工作。这可能不适用于这样的平台：
char'被填充到16或32位边界。
以下文章应该有所帮助：
http://www.eventhelix.com/RealtimeMa...ndOrdering。 htm
-
EventStudio系统设计器2.5 - http://www.EventHelix.com/EventStudio
基于序列图的实时和嵌入式系统设计工具
This type of access is not recommended. In compiler is not adding any
pads so your test program works. This may not work on a platform where
the char''s get padded to 16 or 32 bit boundaries.
The following article should help:
http://www.eventhelix.com/RealtimeMa...ndOrdering.htm
--
EventStudio System Designer 2.5 - http://www.EventHelix.com/EventStudio
Sequence Diagram Based Real-time and Embedded System Design Tool
>我可以对
>Can I use pointer arithmetic on the members of a structure in the
>中的结构成员使用指针算法吗？
>following way?
No.
No.
>我应该担心结构填充吗？
>Should I be worried about structure padding?
是的。
Yes.
>这在我的调试器中工作
但我想知道我是否我在这里弯曲一些规则。
>This works
in my debugger but I wonder if I''m bending some rule here.
正在吹嘘用C4考虑弯曲的smithereens？
Is blowing to smithereens with C4 considered "bending"?
> struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
>struct char_struct {
unsigned char a;
unsigned char b;
unsigned char c;
unsigned char d;
};
如果您想将数组作为结构成员，请将数组用作
结构成员。
If you want an array as a structure member, USE an array as a
structure member.
> struct char_struct test = {50,75,100,225};
unsigned char * c;
c =& test;
如果（c [2] == 100）//这没关系？
>struct char_struct test = {50,75,100,225};
unsigned char *c;
c = &test;
if (c[2]==100) // this is okay?
No.
No.
printf（" c [2] is 100）; //这确实有用
printf("c[2] is 100"); // this does work
在您的YOUR编译器版本上，也许只要你这么做
就不要加一个双结构的第五个成员。
On YOUR version of YOUR compiler and perhaps only as long as you
don''t add a double as a fifth member of the structure.
>
c ++; //这没关系？
>
c++; // this is okay?
是的。将单个变量视为一个元素的数组。你允许
计算一个超过最后一个元素的地址
但不能取消引用它。
Yes. Consider a single variable as an array of one element. You
are allowed to compute the address of "one past the last element"
but not dereference it.
> printf（" * c is 75"）; //这确实有效
>printf("*c is 75"); // this does work
好了，打印一个常量字符串是允许的，但是如果你计划在增加它后取消引用c，那么
，不，那是不允许的。
Gordon L. Burditt
Ok, printing a constant string is allowed, but if you were planning
on dereferencing c after incrementing it, NO, that''s not allowed.
Gordon L. Burditt
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