关于类型转换的问题
我对打字有疑问。这只是此处显示的一个虚拟程序。实际代码太大，无法发布。
typedef struct abc
{
int a;
}abc_t;
main()
{
abc_t *MY_str;
char *p;
MY_str = (abc_t *)p;
}
每当我运行质量分析检查工具时，我都会收到 2 级警告：
Casting to different object pointer type. REFERENCE - ISO:C90-6.3.4 Cast Operators - Semantics <next> Msg(3:3305) Pointer cast to stricter alignment. <next>
任何人都可以告诉我如何解决此问题吗？
I have questions about typecasting. This is just a dummy program shown here. The actual code is too big to be posted.
typedef struct abc
{
int a;
}abc_t;
main()
{
abc_t *MY_str;
char *p;
MY_str = (abc_t *)p;
}
Whenever I run the quality analysis check tool, I get a level 2 warning:
Casting to different object pointer type. REFERENCE - ISO:C90-6.3.4 Cast Operators - Semantics <next> Msg(3:3305) Pointer cast to stricter alignment. <next>
Can anyone please tell me how to resolve this issue?
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评论（4）
简单 - 您的静态分析工具（顺便说一句？）已确定
char*没有特定的对齐要求（它可以指向内存中的任何位置），而abc_t*> 可能有字对齐要求（int必须位于 4/8 字节边界上）。实际上，由于
char*位于堆栈上，因此它在大多数体系结构上都是字对齐的。您的工具看不到这一点。Simple - your static analysis tool (which, btw?) has decided that a
char*does not have a particular alignment requirement (it could point anywhere in memory) whereas anabc_t*likely has a word alignment requirement (intmust be on a 4/8 byte boundary).In reality, as the
char*is on the stack, it will be word aligned on most architectures. Your tool cannot see this.在您的实现（可能还有许多其他实现）中，每个
int必须位于可被sizeof int整除的地址，该地址通常为 4。另一方面，
>char可以位于任何地址。这就像将
3.25分配给int变量。这也是不可能的。因此，当您有一个坏指针时，您的机器可能会出现异常，并且从技术上讲，此代码会调用未定义的行为。
In your implementation (and probably many others) each
intmust be at an address that is divisible bysizeof int, which is often 4.On the other hand, a
charcan be at any address.It's like assigning
3.25to anintvariable. That's also not possible.So when you have a bad pointer, you will probably get an exception from your machine, and technically this code invokes undefined behavior.
char*可以在任何字节边界上对齐，这意味着如果将其转换为结构，则可能无法满足该结构的对齐要求（例如 SIMD 类型所需的 16 字节边界）。a
char*can be aligned on any byte boundary, which means if you cast it to a structure, the alignment requirements of that struct might not be met (such as 16 byte boundaries required for SIMD types).你的代码是无效的C。如果你发现自己在做这样的事情，这可能是更大的误解的结果。例如，我猜测您想从文件/套接字/等读取
abc_t对象。并且您习惯于将char指针传递给read/recv/whatever 函数。相反，您应该只声明一个abc_t类型的对象，并将其地址传递给您正在使用的任何读取函数。Your code is invalid C. If you find yourself doing something like this, it's probably the result of a greater misunderstanding. For instance I'm guessing you want to read an
abc_tobject from a file/socket/etc. and you're used to passing acharpointer to theread/recv/whatever function. Instead you should just declare an object of typeabc_tand pass its address to whatever reading function you're using.