c - 这是严格的别名违规吗？

Question

struct __packed element {
    char val;
};

void foo(struct element arr[4], uint32_t a) {
    uint32_t* i = (uint32_t*)arr;
    *i = a;
}

int main(void) {
    struct element arr[4] __aligned(4);

    foo(arr, 5);
    ...
}

与标题差不多，这是 C 中的严格别名违规吗？

假设arr的存储类型是struct element[4]

Answer 1

是的，这 (*i = a) 是一个严格的别名冲突。

N1570 §6.5 p7:

An object shall have its stored value accessed only by an lvalue expression that has one of the following types: 88)

• a type compatible with the effective type of the object,
• a qualified version of a type compatible with the effective type of the object,
• a type that is the signed or unsigned type corresponding to the effective type of the object,
• a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,
• an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
• a character type.

以上要求都不满足:

• uint32_t 与有效类型 char 不兼容。

• 没有使用限定类型的char。

• uint32_t 不是 char 的无符号版本。

• struct element 没有 uint32_t 成员。

• uint32_t 不是字符类型。

如果原始数组的有效类型为 uint32_t 或使用 malloc 分配，有效类型发生在赋值时，这将是合法的。

uint32_t arr;
foo((struct element*)&arr, 5); // Possible pointer conversion issues still apply

或

void * arr = malloc(4);
foo(arr, 5);

---

注意 uint32_t* i = (uint32_t*)arr 也可能导致未定义的行为，如果转换后的地址不能存储在 uint32_t* 类型变量。但这是特定于实现的，因此取决于您的平台。