C++ 如何将简单指针转换为固定大小的多维数组？

How to cast simple pointer to a multidimensional-array of fixed size?

提问by NoahR

I have a function that takes a pointer to a floating point array. Based on other conditions, I know that pointer is actually pointing to a 2x2 OR 3x3 matrix. (in fact the memory was initially allocated as such, e.g. float M[2][2] ) The important thing is I want to make this determination in the function body, not as the function argument.

我有一个函数，它接受一个指向浮点数组的指针。基于其他条件，我知道指针实际上指向一个 2x2 OR 3x3 矩阵。（实际上内存最初是这样分配的，例如 float M[2][2] ）重要的是我想在函数体中做出这个决定，而不是作为函数参数。

void calcMatrix( int face, float * matrixReturnAsArray )
{
    // Here, I would much rather work in natural matrix notation
    if( is2x2 )
    {
        // ### cast matrixReturnAsArray to somethingAsMatrix[2][2]
        somethingAsMatrix[0][1] = 2.002;
        // etc..
    }
    else if(is3x3)
    { //etc...
    }

}

I am aware that I could use templates and other techniques to better address this problem. My question is really about how to make such a cast at the ### comment. Working in C++.

我知道我可以使用模板和其他技术来更好地解决这个问题。我的问题实际上是关于如何在 ### 评论中进行这样的演员表。在 C++ 中工作。

回答by ecatmur

float (*somethingAsMatrix)[2] = (float (*)[2]) matrixReturnAsArray;

回答by bames53

float *could point to the first element of an array of floats, and ought to be reinterpret_castable to that array type. And the result of that cast could point to the first element of a float [][]and so should be reinterpret_castable to that type, and so on. You ought to be able to compose such casts and just directly do

float *可以指向浮点数组的第一个元素，并且应该 reinterpret_castable 到该数组类型。并且该转换的结果可能指向 a 的第一个元素，float [][]因此应该 reinterpret_castable 到该类型，依此类推。你应该能够组成这样的演员表，直接做

float (&arr)[2][2] = *reinterpret_cast<float (*)[2][2]>(matrixReturnAsArray);

An argument of the type float **is not the same and should not be used this way.

类型的参数float **不同，不应以这种方式使用。

To avoid undefined behavior the pointer must originate from an actual multi-dimensional array, and if the float*is used directly you cannot access more than the first row of the multi-dimensional matrix.

为了避免未定义的行为，指针必须来自实际的多维数组，如果float*直接使用，则不能访问多维矩阵的第一行。

void foo(float *f) {
    f[3] = 10.;

    float (&arr)[2][2] = *reinterpret_cast<float (*)[2][2]>(f);
    arr[1][1] = 10.;
}

void main() {
    float a[2][2];
    foo(&a[0][0]); // f[3] = 10.; is undefined behavior, arr[1][1] = 10. is well defined

    float b[4];
    foo(&b[0]); // f[3] = 10.; is well-defined behavior, arr[1][1] = 10. is undefined
}

Given float arr[2][2];nothing guarantees that &arr[0][1] + 1is the same as &arr[1][0], as far as I have been able to determine. So although you can use a single dimensional array as a multi-dimensional array by doing f[i*width + j]you cannot treat a multi-dimensional array like a single dimensional array.

鉴于float arr[2][2];没有任何保证&arr[0][1] + 1与 相同&arr[1][0]，据我所知。因此，尽管您可以通过这样做f[i*width + j]将一维数组用作多维数组，但您不能将多维数组视为一维数组。

It's better to use C++'s compile-time type-safety instead of just relying on not accidentally passing the wrong thing or performing the wrong reinterpret_cast. To get type-safety using raw-arrays you should use references to the raw array type you want:

最好使用 C++ 的编译时类型安全性，而不是仅仅依靠不意外地传递错误的东西或执行错误的 reinterpret_cast。要使用原始数组获得类型安全，您应该使用对所需原始数组类型的引用：

void foo(float (&f)[2][2]) {}
void foo(float (&f)[3][3]) {}

If you want to pass arrays by value you can't use raw arrays and should instead use something like std::array:

如果要按值传递数组，则不能使用原始数组，而应使用 std::array 之类的内容：

void foo(std::array<std::array<float,2>,2> f) {}
void foo(std::array<std::array<float,3>,3> f) {}

回答by Gort the Robot

This sort of casting is always cleaner, and easier to deal with, with a judicious use of typedef:

这种类型的转换总是更干净，更容易处理，明智地使用 typedef：

typedef float Matrix_t[2][2];

Matrix_t* someThingAsMatrix = (Matrix_t*) matrixReturnAsArray;

If this is C++ and not C, though, you should create a matrix class. (Or better yet, look for an open source one.)

但是，如果这是 C++ 而不是 C，则您应该创建一个矩阵类。（或者更好的是，寻找一个开源的。）