假设我有两个不同的类,它们都以相同的内部方式表示二维坐标数据,如下所示:
Let's say I have two different classes, both represent 2D coordinate data in the same internal way like the following:
class LibA_Vertex{
public:
// ... constructors and various methods, operator overloads
float x, y
};
class LibB_Vertex{
public:
// ... same usage and internal data as LibA, but with different methods
float x, y
};
void foobar(){
LibA_Vertex * verticesA = new LibA_Vertex[1000];
verticesA[50].y = 9;
LibB_Vertex * verticesB = reinterpret_cast<LibB_Vertex*>( vertexA );
print(verticesB[50].y); // should output a "9"
};
鉴于这两个类和上面的函数是相同的,我能否可靠地指望这种指针转换在每种情况下都能按预期工作?
Given the two classes being identical and the function above, can I reliably count on this pointer conversion working as expected in every case?
(背景是,我需要一种简单的方法在两个具有相同 Vertex 类的独立库之间交换顶点数组,并且我想避免不必要地复制数组).
(The background, is that I need an easy way of trading vertex arrays between two separate libraries that have identical Vertex classes, and I want to avoid needlessly copying arrays).
C++11 添加了一个名为 layout-compatible 的概念,适用于此处.
C++11 added a concept called layout-compatible which applies here.
如果两个标准布局结构(第 9 条)类型具有相同数量的非静态数据成员和相应的非静态数据成员,则它们是布局兼容(按声明顺序)具有布局兼容类型(3.9).
Two standard-layout struct (Clause 9) types are layout-compatible if they have the same number of non-static data members and corresponding non-static data members (in declaration order) have layout-compatible types (3.9).
哪里
标准布局类是这样一个类:
A standard-layout class is a class that:
standard-layout struct 是使用 class-key struct
定义的 standard-layout class 或class-key class
.
A standard-layout struct is a standard-layout class defined with the class-key struct
or the class-key class
.
standard-layout union 是使用 class-key union
定义的 standard-layout 类.
A standard-layout union is a standard-layout class defined with the class-key union
.
终于
指向 cv-qualified 和 cv-unqualified 版本 (3.9.3) 的布局兼容的指针类型应具有相同的值表示和对齐要求(3.11).
Pointers to cv-qualified and cv-unqualified versions (3.9.3) of layout-compatible types shall have the same value representation and alignment requirements (3.11).
这保证了 reinterpret_cast
可以将指向一种类型的指针转换为指向任何布局兼容类型的指针.
Which guarantees that reinterpret_cast
can turn a pointer to one type into a pointer to any layout-compatible type.
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