使用映射到 1D 的 2D 数组的访问冲突读取位置

Question

我正在尝试检查数组中的邻居并且没有边缘情况检查限制，程序将导致异常。我至少需要检查bottomLeft、bottomRight、topLeft topRight 角。我在 DirectXTDK 工作，该功能用于平滑景观。

bool Terrain::SmoothenHeightMap(ID3D11Device* device)
{
    bool result;

    int index, nx, nz;
    float height = 0.0;
    int neighbours[8] = {}; // array starts at 0, inclusive
    int n = 8;

    /* Initialise corner of height map */ // 1.
    int bottomLeftCorner = 0;
    int bottomRightCorner = (m_terrainHeight * (m_terrainHeight - 1));
    int topLeftCorner = (m_terrainWidth - 1);
    int topRightCorner = m_terrainHeight * (m_terrainHeight - 1) + (m_terrainWidth - 1);

    m_frequency = (6.283 / m_terrainHeight) / m_wavelength; //we want a wavelength of 1 to be a single wave over the whole terrain.  A single wave is 2 pi which is about 6.283
    // m_terrainHeight is actually the z axis

    for (int j = 0; j < m_terrainHeight; j++)
    {
        for (int i = 0; i < m_terrainWidth; i++)
        {
            index = (m_terrainHeight * j) + i; 

            float sum = m_heightMap[index].y;

            // with more than 128 square dimensions, initial neighbours on bottom row might not exist
            // can refractor this better if it works

            if (m_heightMap[(m_terrainHeight * (j - 1)) + (i - 1)].x != NULL) {
                neighbours[0] = m_heightMap[(m_terrainHeight * (j + 1)) + (i - 1)].y; // top left
                neighbours[1] = m_heightMap[(m_terrainHeight * (j + 1)) + (i)].y;     // top middle
                neighbours[2] = m_heightMap[(m_terrainHeight * (j + 1)) + (i + 1)].y; // top right
                neighbours[3] = m_heightMap[(m_terrainHeight * (j)) + (i - 1)].y;     // middle left
                neighbours[4] = m_heightMap[(m_terrainHeight * (j)) + (i + 1)].y;     // middle right
                neighbours[5] = m_heightMap[(m_terrainHeight * (j - 1)) + (i - 1)].y; // bottom left
                neighbours[6] = m_heightMap[(m_terrainHeight * (j - 1)) + (i)].y;     // bottom middle
                neighbours[7] = m_heightMap[(m_terrainHeight * (j - 1)) + (i + 1)].y; // bottom right
            }




            for (int z = 0; z < n; z++)
            {
                if (neighbours[z] < 0 || neighbours[z] >= m_terrainHeight * m_terrainWidth) // if out of map, take y of current index for sum
                {
                    sum += m_heightMap[index].y;
                }
                else
                {
                    sum += neighbours[z]; // if exists, include in sum
                }
            }

            // smoothen based on neighbours
            m_heightMap[index].y = sum / 9.0f; // current point n is no. of neighbours +1 for current vertex point// total of 9 points in a 3*3 grid
        }
    }


    result = CalculateNormals();
    if (!result)
    {
        return false;
    }

    result = InitializeBuffers(device);
    if (!result)
    {
        return false;
    }
}

上述代码用于平滑地形，使用高度图通过存储在结构中的点。以下是结构体：

struct HeightMapType
    {
        float x, y, z;
        float nx, ny, nz;
        float u, v;
    };

通过创建一个名为m_heightmap的指针来使用它。

HeightMapType* m_heightMap;

Answer 1

由于我没有完整的代码块，我可以告诉你找到错误的可能方法。

正如我在屏幕截图中看到的，错误行不能喜欢与数组关联的值或位置。我建议您调试代码并单步执行每个 for 循环以获得 m_terrainHeight * (j-1) + (i-1) 的执行值。 可能，上面的表达式正在评估一个意外的值。

对所有分配和执行重复相同的逐行执行过程，以检查您是否获得了预期值。

请分享整个代码块，以便更好地理解问题。

Answer 2

我设法通过以下方式检查角落来解决它：

if (m_heightMap[index].z + 1 >= m_terrainHeight || m_heightMap[index].z <= 0) {
                // if no neighbours, set height to that of current index/element
                neighbours[0], neighbours [1], neighbours[2] = m_heightMap[index].y;
                }
            else {
                neighbours[0] = m_heightMap[(m_terrainHeight * (j + 1)) + (i - 1)].y; // top left
                neighbours[1] = m_heightMap[(m_terrainHeight * (j + 1)) + (i)].y;     // top middle
                neighbours[2] = m_heightMap[(m_terrainHeight * (j + 1)) + (i + 1)].y; // top right
            }
            neighbours[3] = m_heightMap[(m_terrainHeight * (j)) + (i - 1)].y;     // middle left
            neighbours[4] = m_heightMap[(m_terrainHeight * (j)) + (i + 1)].y;     // middle right
            if (m_heightMap[index].z <= 0 || m_heightMap[index].z + 1 >= m_terrainHeight) {
                // if no neighbours, set height to that of current index/element
                neighbours[5], neighbours[6], neighbours[7] = m_heightMap[index].y;
            }
            else {
                neighbours[5] = m_heightMap[(m_terrainHeight * (j - 1)) + (i - 1)].y; // bottom left
                neighbours[6] = m_heightMap[(m_terrainHeight * (j - 1)) + (i)].y;     // bottom middle
                neighbours[7] = m_heightMap[(m_terrainHeight * (j - 1)) + (i + 1)].y; // bottom right
            }