本文介绍了初始化未知维度数组的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述 29岁程序员，3月因学历无情被辞！ 我很惊讶，我找不到这个问题。I was surprised that I couldn't find this question existing. I've tried to generalize it (with some nice untested code) to something everyone can benefit from.假设我有一个多维的点Suppose I have a multidimensional Point:template <int dims> class Point { public: double data[dims]; };现在我创建了一个多维数组：Now I create a multidimensional array of them: template <int dims> void foobar(int count0, ...) { //Using variadic function. Could also use variadic templates in C++ (arguably better) int counts[dims], total_count=count0; counts[0]=count0; va_list args; va_start(args,count0); for (int i=1;i<dims;++i) { int count = va_arg(args,int); counts[i] = count; total_count *= count; } va_end(args); Point<dims>* array = new Point<dims>[total_count]; //...}请参阅数组是一个未知维度的多维数组，以1D数组表示。As you can see, array is a multidimensional array of unknown dimensionality, represented in a 1D array. 我的问题：如何清洁地将这个数组初始化到多维网格点？这里是我想要的1，2和3维的示例行为。显然，我不想为每一个可能的维度写我想使用！目标是概括。Here's the example behavior I want in 1, 2, and 3 dimensions. Obviously, I don't want to write this for every possible dimensionality I might want to use! The goal is to generalize this.//Example: dim==1for (int x=0; x<counts[0]; ++x) { Point<1>& point = array[x]; point.data[0] = (x+0.5) / (double)counts[0];}//Example: dim==2for (int y=0; y<counts[1]; ++y) { for (int x=0; x<counts[0]; ++x) { Point<2>& point = array[y*counts[0]+x]; point.data[0] = (x+0.5) / (double)counts[0]; point.data[1] = (y+0.5) / (double)counts[1]; }}//Example: dim==3for (int z=0; z<counts[2]; ++z) { for (int y=0; y<counts[1]; ++y) { for (int x=0; x<counts[0]; ++x) { Point<3>& point = array[(z*counts[1]+y)*counts[0]+x]; point.data[0] = (x+0.5) / (double)counts[0]; point.data[1] = (y+0.5) / (double)counts[1]; point.data[2] = (z+0.5) / (double)counts[2]; } }}注意：我想出了一些讨厌的方法，并且他们不愉快和缓慢。特别是，我想避免递归，如果可能，因为这将被调用高维小型数据集很频繁。 注意：在C中有明显的并行，所以C或C ++都很好。 C ++ 11首选。Note: I've come up with a few nasty ways, and they're inelegant and slow. Especially, I want to avoid recursion, if possible, since this will be called on high-dimensional smallish datasets quite frequently.Note: There are obvious parallels in C, so either C or C++ is fine. C++11 preferred.推荐答案从 X，Y，Z 到扁平数组（F），我们有以下等式：Going from X,Y,Z to the flattened array (F) we have the following equationF=(Z*DimY+y)*DimX+X或F=Z*DimY*DimX+Y*DimX+XX = F % DimXY = F % DimX*DimY/DimXZ = F % DimX*DimY*DimZ/DimX*DimY 3 * 3 * 5 + 1 * 5 + 2 = 45 + 5 + 2 = 52in a 7 x 3 x 5 array, Z=3, Y=1, X=2 would be at 3*3*5 + 1*5 + 2= 45+5+2=52X = `52` % 5 = 2Y = `52` % (5 * 3) / 5 = 7 / 5 = 1Z = `52` % (7 * 5 * 3)/15 = 52/15 = 3 数组，Z = 4，Y = 2，X = 3将在 4 * 3 * 5 + 2 * 5 + 3 = 60 + 10 + 3 = 73in a 7 x 3 x 5 array, Z=4, Y=2, X=3 would be at 4*3*5 + 2*5 + 3= 60+10+3=73X = `73` % 5 = 3Y = `73` % (5 * 3) / 5 = 13 / 5 = 2Z = `73` % (7 * 5 * 3)/15 = 73/15 = 4数组中的累积乘积， mult {1，X，X * Y，X * Y * Z，...} 和数组中的所有点， valIf we keep the cumulative products in an array, mult { 1, X, X*Y, X*Y*Z, ...} and all points in an array, val指向平面数组：F=sum(mult[i]*val[i]);平面数组到点：i[0]=F%mult[1]/mult[0];i[1]=F%mult[2]/mult[1];...然后我们可以遍历F从索引到平面数组所有点：X，Y，...如上所述，并在通用循环中进行初始化：We can then iterate over F (the flat array) , reverse engineer from the index into the flat array all points: X,Y,... as above and do the initialization you want in a generic loop:给定 mult as mult [0] = 1; mult [d + 1] = mult [d] * count [d];for (int i = 0; i < total_count; ++i) { for (int d=0; d < dims; ++d) { int dim=(i%mult[d+1])/mult[d]; point.data[d] = (dim+0.5) / (double)counts[d]; }} 这篇关于初始化未知维度数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！ 上岸，阿里云！
