C中的结构扩展 [英] Struct Extension in C

问题描述

假设定义了2种结构的广告，例如：

Suppose there are 2 structures defined ad such:

typedef struct {
   T x;
   T y;
} A;

typedef struct {
   A a;
   T z;
} B;

我可以将指向结构B的指针视为指向结构A的指针吗？

Can I treat a pointer to structure B as pointer to structure A?

实际上，这个可靠/标准/便携式/编译器不变的变量是：

In practice is this reliable/standard/portable/compiler-invariant:

B b = {{1,2},3};
A * a = &b;

print(a->x);
print(a->y);

推荐答案

C17 6.7.2.1指出了这一点（强调我的意思）：

C17 6.7.2.1 states this (emphasis mine):

在结构对象中，非位字段成员和位字段
所在的单元的地址按顺序增加在其中声明它们。 经过适当转换的指向
结构对象的指针指向其初始成员（或者，如果该成员是
位域，则指向它所在的单元） ，，反之亦然。

Within a structure object, the non-bit-field members and the units in which bit-fields reside have addresses that increase in the order in which they are declared. A pointer to a structure object, suitably converted, points to its initial member (or if that member is a bit-field, then to the unit in which it resides), and vice versa.

这意味着您必须适当地转换 B b 对象的指针指向其第一个成员的类型。此转换不是隐式发生的，您必须通过显式强制转换来做到这一点：

This means that you must "suitably convert" the pointer for the B b object into the type of it's first member. This conversion doesn't happen implicitly, you must do this by means of an explicit cast:

A * a = (A*)&b;

按照上面引用的部分，这样做是明确且安全的。

Doing so is well-defined and safe, as per the quoted part above.

类似地，不允许编译器假定指向 A 的指针和指向 B 的指针别名。有效类型6.5.7的规则（严格别名）在这种情况下给出了例外：

Similarly, the compiler is not allowed to assume that a pointer to A and a pointer to B don't alias. The rule of effective type 6.5.7 ("strict aliasing") gives an exception for this case:

对象只能通过以下方式访问其存储值：具有以下类型之一的左值表达式：

...

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

• 包括以下类型之一的聚合或并集类型

例如，在优化过程中，编译器调用函数 void func （B * b）不允许假定外部线性变量 extern A a; 在其他翻译单元中定义的值未被更改功能。

For example, during optimization a compiler calling the function void func (B* b) is not allowed to assume that a external linage variable extern A a; defined in some other translation unit was not changed by the function.

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