C / C ++结构体内存布局当量 [英] C/C++ Struct memory layout equivalency

问题描述

考虑下面的C结构和C ++的struct声明：

Consider the following C struct and C++ struct declarations:

extern "C" { // if this matters
typedef struct Rect1 {
  int x, y;
  int w, h;
} Rect1;
}

struct Vector {
  int x;
  int y;
}

struct Rect2 {
  Vector pos;
  Vector size;
}

• 是 Rect1的和 RECT2 对象始终相同？

  具体来说，我可以放心地 reinter pret_cast 从 RECT2 * 到 Rect1的* 并假设在 RECT2 对象全部四个 INT 值匹配一对一的四个 INT S IN Rect1的？

  Specifically, can I safely reinterpret_cast from Rect2* to Rect1* and assume that all four int values in the Rect2 object are matched one on one to the four ints in Rect1?

  这是否有所作为，如果我换 RECT2 到非POD类型，例如通过增加一个构造函数？

  Does it make a difference if I change Rect2 to a non-POD type, e.g. by adding a constructor?

  推荐答案

  
  
  • 我会这样想，但我也认为有可能（合法）为 RECT2 :: POS之间填充和 RECT2 ::大小。因此，要确保，我想补充编译器特定的属性为包的领域，从而保证所有的 INT 是相邻的，紧凑。这是不太关于C与C ++，更多的是，你有可能使用两个不同的编译器在两种语言进行编译时，即使这些编译器来自单一供应商的事实。
  
  
  • 使用 reinter pret_cast 的指针转换为一种类型的指针到另一个，你很可能违反严格走样的规则。假设你解引用指针之后，你们会在这种情况下。
  
  
  • 添加一个构造函数不会改变布局（虽然它会使类非POD），但添加访问修饰符如私人之间的两个领域可能会改变布局（在实践中，不仅在理论）。
  
  
  • I would think so, but I also think there could (legally) be padding between Rect2::pos and Rect2::size. So to make sure, I would add compiler-specific attributes to "pack" the fields, thereby guaranteeing all the ints are adjacent and compact. This is less about C vs. C++ and more about the fact that you are likely using two "different" compilers when compiling in the two languages, even if those compilers come from a single vendor.
  • Using reinterpret_cast to convert a pointer to one type to a pointer to another, you are likely to violate "strict aliasing" rules. Assuming you do dereference the pointer afterward, which you would in this case.
  • Adding a constructor will not change the layout (though it will make the class non-POD), but adding access specifiers like private between the two fields may change the layout (in practice, not only in theory).

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