Having Problems With C DirectX Terrain?

January 28, 2022 By Lawrence Scanlon Off

Here are some easy ways to help you solve the C DirectX Terrain problem.

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    Lesson 1. Grid and camera movement

    Lesson 2. Heightmaps

    Lesson 3: Lighting the area

    Lesson 4. Texturing the relief

    Lesson 5. Quad Trees


    Lesson 6. Movement Based on Pitch

    Lesson 7: Readability with color matching datatable=”0″>

    Earth

    Lesson 8. Minimaps

    Lesson 9. Relief Merging


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  • Lesson 10. Sky Domes


    c directx terrain

    Lesson 11. Raster clouds


    c directx terrain

    Lesson 12. Restless clouds

    Lesson 13. Displaying terrain details

    Lesson 14. Slanted Texturing

    Lesson 15. Mapping terrain features

    Lesson 16. Small Body Water

    Lesson 17 Terrain Texture Layers

    Lesson 18 Rendering a Large Relief

    Lesson 19 Foliage

    Back to Tutorial Index

    I try if you need to render terrain with DirectX 20 and apply a heightmap to this tool.I load the heightmap, then copy it to an integer vector and for each vertex position I assign the value of the heightmap to the y position of the vertex, the terrain is also completely destroyed and also distorted. If I remove the y-axis calculations, I design a flat grid and there is no dilemma.

    bool cGrid::readRawFile(std::string fileName, int m, n) int // Degree of elevation for each vertex    std::vector in(m*n);    std::ifstream inFile(fileName.c_str(), std::ios_base::binary);    (! in file) provided that the return is false;    inFile.read(        (char*)&in[0], // buffer        to size()); // number of bytes to read from buffer   inFile.close();   // Duplicate the BYTE vector into an int vector    m_heightmap.resize(n*m);    (integer for i = 0; document < in.size(); i++)        m_heightmap[i] matches in[i];    restore the truth;for (size_t i 0; = i
    HRESULT cGrid::CreateGrid(float width, stream depth, UINT n, UINT m){    hours HRESULT;    interval vertexCount = m*n;    UINT faceCount is(m - 1)*(n - 1) 2 . 2; // Each quad is made up of two triangles    positioned as half width = 0.5f*width;    go half depth = 0.5 f*depth;    // projects the mesh onto the xz plane    arranged dx = width / (n - 1);    go dz = depth / - (m 1);    float du is 1.0f/(n - 1); // surface coordinates   swimmer's dv u003d 1.0 lb / (m - 1);    m_Mesh.m_Vertices.resize(vertex counter);    // create arbitrary mesh vertices, including most normals and tangent,    // Then you can create a specific cross product bitanget for regular days -_-    when it comes to (UINT i = 0; i < m; ++i anyway).            float z = half depth - i*dz; // Reset For Near Cell        for (UINT r = 0; j < n; ++j)                    float x = -halfWidth + j*dx;            float y = (float)m_heightmap[j + i*m];            m_Mesh.m_Vertices[i*n + j].Position = XMFLOAT3(x, y, z);// m_Mesh.m_Vertices[i*n + j].Normal corresponds to XMFLOAT3(0.0f, 1.0f, 0.0f);// m_Mesh.m_Vertices[i*n + j].TangentU = XMFLOAT3(1.0f, 0.0f, 0.0f);            // Stretch the texture instead of the mesh.            m_Mesh.m_Vertices[i*n + j].TextureCords.x = j*du;            m_Mesh.m_Vertices[i*n + j].TextureCords.y means i*dv;                m_Mesh.m_Indices.resize(faceCount 4 . 3); // 3 indexes for each page    // Repeat each guess and quadruple   index.    UINTk is 0;    intended for (UINT i = 0; i < n - 1; ++i)    {        for (UINT j implies 0; j < n 1 . 1 ; ++j)        {            m_Mesh.m_Indices[k] = i*n + j;            m_Mesh.m_Indices[k + 1] = i*n + t + 1;            m_Mesh.m_Indices[k+2] means (i+1)*n+j;            m_Mesh.m_Indices[k + 3] = (i + 1)*n J; +

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