内部结构中的尾部填充

Question

编译器通常在结构上插入尾部填充以满足对齐在数组中使用时的限制：

// Size 4, alignment 2
struct A {
    uint16_t x;
    uint8_t y;
    // Padding 1 byte
};

// Size 6, alignment 2
struct B {
    struct A xy;
    uint8_t z;
    // Padding 1 byte
};

现在考虑使用内部结构组合这些：

struct AB {
    struct {
        uint16_t x;
        uint8_t y;
    } xy;
    uint8_t z;
};

以下可以容纳 4 个字节，同时不违反对齐限制。此外，内部结构 xy 没有可以在其他地方使用的类型，所以不需要尾随填充。

缺点是成员 xy 与 struct A 不兼容，但有没有理由应该像它们在不同类型定义中那样。

是否允许编译器进行这种大小优化？

或者，换句话说，标准是否要求 2 个具有相同成员的结构将始终导致相同的布局？

Answer 1

答案很可能由兼容类型的概念给出。如果 t1 和 t2 是兼容类型，则指向 t1 的指针可用于访问使用类型 t2 设置的内存。

在 C11 标准中：

6.2.7 Compatible type and composite type

… Moreover, two structure, union, or enumerated types declared in separate translation units are compatible if their tags and members satisfy the following requirements: If one is declared with a tag, the other shall be declared with the same tag. If both are completed anywhere within their respective translation units, then the following additional requirements apply: there shall be a one-to-one correspondence between their members such that each pair of corresponding members are declared with compatible types; if one member of the pair is declared with an alignment specifier, the other is declared with an equivalent alignment specifier; and if one member of the pair is declared with a name, the other is declared with the same name. …

两个没有相同标签的结构是不兼容的类型，我没有看到任何强制它们具有相同布局的东西。

内部结构中的尾部填充

Trailing padding in internal structure

c

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