将double数组转换为double的结构 [英] Casting double array to a struct of doubles

问题描述

可以将双精度数组强制转换为由双精度组成的结构吗?

Is it OK to cast a double array to a struct made of doubles?

struct A
{
   double x;
   double y;
   double z;
};

int main (int argc , char ** argv)
{
   double arr[3] = {1.0,2.0,3.0};
   A* a = static_cast<A*>(static_cast<void*>(arr));
   std::cout << a->x << " " << a->y << " " << a->z << "\n";
}

这将打印1 2 3.但是是否可以保证每次都可以与任何编译器一起工作?

This prints 1 2 3. But is it guaranteed to work every time with any compiler?

根据

9.2.21:指向标准布局结构对象的指针，已适当转换?使用reinterpret_cast指向其初始成员(...)，反之亦然.

9.2.21: A pointer to a standard-layout struct object, suitably converted ? using a reinterpret_cast, points to its initial member (...) and vice versa.

如果我将代码替换为

struct A
{
  double & x() { return data[0]; }
  double & y() { return data[1]; }
  double & z() { return data[2]; }
private:
   double data[3];
};

int main (int, char **)
{
   double arr[3] = {1.0,2.0,3.0};
   A* a = reinterpret_cast<A*>(arr);
   std::cout << a->x() << " " << a->y() << " " << a->z() << "\n";
}

然后保证它可以工作.正确的?我知道很多人不会觉得这令人愉悦，但是使用结构而不需要复制输入数组数据有很多好处.我可以在该结构中定义成员函数，以计算标量和向量乘积，距离等，这将使我的代码比使用数组更容易理解.

then it is guaranteed to work. Correct? I understand that many people would not find this aesteticaly pleasing but there are advantages in working with a struct and not having to copy the input array data. I can define member functions in that struct to compute scalar and vector products, distances etc, that will make my code much easier to understand than if I work with arrays.

怎么样

int main (int, char **)
{
   double arr[6] = {1.0,2.0,3.0,4.0,5.0,6.0};
   A* a = reinterpret_cast<A*>(arr);
   std::cout << a[0].x() << " " << a[0].y() << " " << a[0].z() << "\n";
   std::cout << a[1].x() << " " << a[1].y() << " " << a[1].z() << "\n";
}

这是否还可以保证正常工作，或者编译器可以在数据成员之后放置一些内容，以便sizeof(A) > 3*sizeof(double)?有没有可移植的方法来阻止编译器这样做?

Is this also guaranteed to work or the compiler could put something AFTER the data members so that sizeof(A) > 3*sizeof(double)? And is there any portable way to prevent the compiler from doing so?

推荐答案

不，不能保证.

唯一常识是禁止任何编译器在x和y之间或在y和z之间插入填充.任何语言标准中都没有禁止它的规则.

The only thing prohibiting any compiler from inserting padding between x and y, or between y and z is common sense. There is no rule in any language standard that would disallow it.

即使没有填充，即使A的表示形式与double[3]的表示形式完全相同，也仍然无效.该语言不允许您假装一种类型确实是另一种类型.甚至不允许您将struct A { int i; };的实例视为struct B { int i; };.

Even if there is no padding, even if the representation of A is exactly the same as that of double[3], then it's still not valid. The language doesn't allow you to pretend one type is really another type. You're not even allowed to treat an instance of struct A { int i; }; as if it's a struct B { int i; };.

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