C-在一维字节数组中索引x，y，z坐标 [英] C- Indexing x,y,z coordinates in a 1D byte array

问题描述

我想评估曲面的值以实施行进的四面体算法，但我不了解如何使用.raw无格式数据。

I want to evaluate the values of the surface to implement a marching tetrahedron algorithm, but I don't understand how to work with .raw unformatted data.

将带有卷数据集的.raw文件加载到一维字节数组中后，应应用哪种算术转换从中获取与X，Y，Z关联的值？这是我知道加载.raw文件的唯一方法，是否可以创建3D字节数组而不是此格式？怎么样？

After loading a .raw file with a volume dataset into a 1D byte array, what arithmetic transformation should be applied to get the value associated to X,Y,Z from it? This is the only way I know to load .raw files, could I create a 3D byte array instead of this? How?

int XDIM=256, YDIM=256, ZDIM=256;
const int size = XDIM*YDIM*ZDIM;
bool LoadVolumeFromFile(const char* fileName) {

    FILE *pFile = fopen(fileName,"rb");
   if(NULL == pFile) {
    return false;
   }

   GLubyte* pVolume=new GLubyte[size]; //<- here pVolume is a 1D byte array 
   fread(pVolume,sizeof(GLubyte),size,pFile);
   fclose(pFile);

推荐答案

在（x ，y，z）是：

The way you'd index the datum at (x, y, z) is:

pVolume[((x * 256) + y) * 256 + z]

在幕后，这是C编译器为您编写的：

Behind the scenes, this is what the C compiler does for you if you write:

GLuByte array[256][256][256];

array[x][y][z]

仅适用那仅仅是因为C从0开始索引；如果从1开始对语言进行索引，则在进行索引之前，必须对计算进行修改以实现通过从x，y和z中减去一个来获得最终结果。

It only works that simply because C indexes from 0; if the language indexed from 1, you'd have to revise the calculation to achieve the net result obtained by subtracting one from each of x, y and z before doing the indexing.

您能概括任意尺寸的公式吗？

Can you generalize the formula for arbitrary dimensions?

给出（其中数值并不重要）：

Given (where the numeric values don't really matter):

DIMx = 256
DIMy = 128
DIMz =  64

一维数组 pData 中（x，y，z）处的基准位于：

the datum at (x, y, z) in 1D array pData is found at:

pData[((x * DIMx) + y) * DIMy + z]

DIMz的值主要用于验证： 0< = z< DIMz （使用数学表示法，而不是C表示法），并且并行 0< = x< DIMx; 0< = y< = DIMy 。 z 的C表示法是 0< = z& & DIMz ;对 x 和 y 重复必要修改。

The value of DIMz serves primarily for validation: 0 <= z < DIMz (using mathematical rather than C notation), and in parallel 0 <= x < DIMx; 0 <= y <= DIMy. The C notation for z is 0 <= z && z < DIMz; repeat mutatis mutandis for x and y.

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