检查 GUID 是否为空(在 C 中)

问题描述

我想检查一个 GUID 结构是否为空/所有字段都是 0.这是我写的代码:

I want to check if a GUID structure is empty/all fields are 0. This is the code I wrote:

#include <windows.h>

static BOOL IsEmptyGuid(const GUID * const pGuid)
{
    return \
    (pGuid->Data1 == 0) &&
    (pGuid->Data2 == 0) &&
    (pGuid->Data3 == 0) &&
#ifdef _WIN64
    (*(DWORD64 *)pGuid->Data4 == 0);
#else
    (*(DWORD *)pGuid->Data4 == 0) && (*(DWORD *)(pGuid->Data4 + 4) == 0);
#endif
}

/* GUID definition from MSDN
typedef struct _GUID {
    DWORD Data1;
    WORD  Data2;
    WORD  Data3;
    BYTE  Data4[8];
} GUID;
*/


int main() {
    GUID guid1;
    guid1.Data1 = 0;
    guid1.Data2 = 0;
    guid1.Data3 = 0;
    memset(guid1.Data4, 0x0, 8);

    printf("Result: %u\n", IsEmptyGuid(&guid1));
}

检查字段 Data4 是否等于 0 的更安全方法是遍历每个字节并检查其值.但是，我发现上面的代码更具表现力.

A safer way to check if field Data4 equals 0 is to iterate over every byte and check for it's value. But, I find the code from above more expressive.

我想知道，对吗?安全吗?

I would like to know, is it correct? Is it safe?

谢谢！

推荐答案

代码不正确.它违反了严格的别名规则(N1570 §6.5 p7)，导致未定义的行为.

Code is incorrect. It breaks strict aliasing rule (N1570 §6.5 p7), causing undefined behaviour.

一个对象只能通过左值表达式访问其存储的值，该表达式具有以下之一以下类型:

• 与对象的有效类型兼容的类型，
• 与对象的有效类型兼容的类型的限定版本，
• 一个类型，它是对应于有效类型的有符号或无符号类型对象，
• 一种类型，它是对应于限定版本的有符号或无符号类型对象的有效类型，
• 一个聚合或联合类型，其中包括上述类型之一成员(包括递归地，子聚合或包含联合的成员)，或
• 一种字符类型.

确切地说，UB 在您使用非匹配类型取消引用指针时发生:

To be exact, UB happens when you dereference pointer using non-matching type:

DWORD64 * temp = (DWORD64 *)pGuid->Data4; // Allowed, but implementation defined
DWORD64 temp2 = *temp;                    // Undefined behaviour

使用循环单独检查每个元素，或与 memcmp 比较到相同大小的零填充数组.

Use loop to check each element individually, or compare with memcmp to zero filled array of the same size.

如评论中所述，某些编译器允许禁用严格别名，但应避免这样做，因为它会降低代码的可移植性，并且您仍然存在潜在的对齐问题.

As noted in the comments, some compilers allow disabling strict aliasing, but that should be avoided as it makes code less portable, and you still have potential alignment issues.

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