嵌套结构和在C严格别名 [英] Nested structs and strict aliasing in c

问题描述

请考虑以下code：

typedef struct {
  int type;
} object_t;

typedef struct {
  object_t object;
  int age;
} person_t;

int age(object_t *object) {
  if (object->type == PERSON) {
    return ((person_t *)object)->age;
  } else {
    return 0;
  }
}

这是法律code，或者它违反了C99严格别名规则？请解释为什么它是合法的/不合法的。

Is this legal code or is it violating the C99 strict aliasing rule? Please explain why it is legal/illegal.

推荐答案

严格别名规则是关于两种不同类型的引用相同的内存位置的（ISO / IEC9899 / TC2）。虽然你的例子reinter $ P $点的object_t对象的地址为 PERSON_T 的地址，它不参考内存在 object_t 位置通过reinter preTED指针，因为年龄位于过去的边界 object_t 。由于通过指针引用的内存地址是不一样的，我会说，这不是违反了严格别名规则。 FWIW， GCC -fstrict走样-Wstrict走样= 2 -O3 -std = C99 似乎与考核同意，并且不产生警告。

Strict aliasing rule is about two pointers of different types referencing the same location in memory (ISO/IEC9899/TC2). Although your example reinterprets the address of object_t object as an address of person_t, it does not reference memory location inside object_t through the reinterpreted pointer, because age is located past the boundary of object_t. Since memory locations referenced through pointers are not the same, I'd say that it is not in violation of the strict aliasing rule. FWIW, gcc -fstrict-aliasing -Wstrict-aliasing=2 -O3 -std=c99 seems to agree with that assessment, and does not produce a warning.

这是不够的，决定了它是合法的code，虽然：你的榜样，使一个假设，即嵌套结构的地址是一样的，它的外部结构。顺便说一下，这是一个安全的假设根据C99标准做出

This is not enough to decide that it's legal code, though: your example makes an assumption that the address of a nested structure is the same as that of its outer structure. Incidentally, this is a safe assumption to make according to the C99 standard:

6.7.2.1-13。一个指向结构对象，适当转换，道出了它的初始成员

6.7.2.1-13. A pointer to a structure object, suitably converted, points to its initial member

以上两方面的考虑，使我想到你的code是合法的。

The two considerations above make me think that your code is legal.

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