我可以在 C 中对 void * 指针进行算术运算吗？
这是有效的
void *p = &X; /* some thing */
p += 12;
吗？如果有效的话，p 现在指向什么？ 我有（第三方）代码可以执行此操作（并且可以干净地编译），我的猜测是 void * 被视为 char *。我值得信赖的 K&R 在这个主题上保持沉默
编辑：我的小测试应用程序在 gcc 4.1.1 上运行良好，并将 void * 视为 char *。但是 g++ barfs
我知道如何正确地做到这一点。我需要知道是否必须清理此代码库才能找到其完成的所有位置。
BTW gcc -pedantic 抛出警告
摘要：
C 规范不明确。它说，就表示形式和用作函数参数而言，void*=char*。但对于指针算术却保持沉默。
- gcc (4) 允许它并将其视为 char *
- g++ 拒绝它
- gcc -pedantic 警告它
- vs2010 c 和 c++ 拒绝它
is this valid
void *p = &X; /* some thing */
p += 12;
and if so what does p now point to?
I have (third party) code that does this (and compiles cleanly) and my guess is that the void * was treated as a char *. My trusty K&R is silent(ish) on the topic
EDIT: My little test app runs fine on gcc 4.1.1 and treats void * as char *. But g++ barfs
I know how to do it properly. I need to know if I have to clean this code base to find all the places its done.
BTW gcc -pedantic throws up a warning
Summary:
The C spec is ambiguous. It says that in terms of representation and use as function parameters void* =char*. But it is silent regarding pointer arithmetic.
- gcc (4) permits it and treats it as
char * - g++ refuses it
- gcc -pedantic warns about it
- vs2010 both c and c++
refuses it
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不，这是不合法的。
void*不能任意递增。它需要首先转换为特定类型。如果您想将其增加特定的字节数，那么这就是我使用的解决方案。
char类型很方便，因为它的定义大小为 1 个字节。编辑
有趣的是它在 gcc 上运行并带有警告。我在 Visual Studio 2010 上进行了测试并验证它无法编译。我对标准的有限理解会说 gcc 在这里出错。您可以添加以下编译标志吗
No this is not legal. A
void*cannot be arbitrarily incremented. It needs to be cast to a specific type first.If you want to increment it by a specific number of bytes then this is the solution I use.
The
chartype is convenient because it has a defined size of 1 byte.EDIT
It's interesting that it runs on gcc with a warning. I tested on Visual Studio 2010 and verified it does not compile. My limited understanding of the standard would say that gcc in the error here. Can you add the following compilation flags
引用规范：
指向 void 的指针不是指向对象类型的指针，根据以下摘录：
因此，对于指向 void 类型的指针未定义指针算术。
To quote from the spec:
A pointer to void is not a pointer to an object type, as per these excerpts:
Therefore, pointer arithmetic is not defined for pointer to void types.
这取决于编译器。那些允许它的人将 sizeof(*(void *)) 视为 1。
编辑：它仅适用于 void 指针算术。在这种情况下，使用 sizeof(int) 或 0 的步长是没有意义的。使用它的人的共同期望是尽可能最小的步长。
It depends on compiler. Those that allow it consider sizeof(*(void *)) as 1.
EDIT: it's only for void pointer arithmetic. It would have no sense using in this case steps of sizeof(int) or of 0. The common expectations of someone who uses it would be the smallest possible step.
你的猜测是正确的。
在标准 ISO C99 的第 6.2.5 节第 26 段中，它声明 void 指针和字符指针将具有相同的表示和对齐要求（释义）。
Your guess is correct.
In the standard ISO C99, section 6.2.5 paragraph 26, it declares that void pointers and character pointers will have the same representation and alignment requirements (paraphrasing).
您可能想查看此问题的最高投票答案
指针算术C 中的空指针
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Pointer arithmetic for void pointer in C
我认为你不能，因为它不知道它的类型，因此无法寻求正确的字节数。
首先将其转换为类型，即
(int)。I don't think you can, because it doesn't know its type, therefore can not seek the correct amount of bytes.
Cast it to a type first, i.e.
(int).