使用iframe做网站,短视频营销推广,室内设计培训,桂林最新新闻1. 背景#xff1a; 项目中使用到了纹理进行插值的加速#xff0c;因此记录一些自己在学习tex2D的一些过程 2. 代码#xff1a; #include cuda_runtime.h #include device_launch_parameters.h #include assert.h #include stdio.h 项目中使用到了纹理进行插值的加速因此记录一些自己在学习tex2D的一些过程 2. 代码 #include cuda_runtime.h #include device_launch_parameters.h #include assert.h #include stdio.h #include iostream #include cuda_fp16.h #include vectorvoid Data2Half(half* pDst, const int16_t* pSrc, const int Ndots); static __global__ void Tex2DTest(cudaTextureObject_t p_rf_data, float* pfRes1, float* pfRes2);static __global__ void data2half(half* pDst, const int16_t* pSrc, const int Ndots) {const int tid blockIdx.x * blockDim.x threadIdx.x;if (tid Ndots)return;pDst[tid] __short2half_rn(pSrc[tid]); }cudaTextureObject_t m_tex 0; cudaArray* m_pRFData nullptr; int16_t* m_i16RFDataBuffer nullptr; // 设备端的RF数据 half* m_pHalfRFDataCache nullptr; // 转换为半浮点型的RF数据缓存用于将SHORT类型转换为FLOAT类型int main() {const int nRx 2;const int Nsample 2;const int IQ 1;cudaError_t error;cudaChannelFormatDesc channelDesc cudaCreateChannelDescHalf();error cudaMallocArray(m_pRFData, channelDesc, nRx * IQ, Nsample, cudaArrayTextureGather);assert(m_pRFData);cudaResourceDesc texRes;memset(texRes, 0, sizeof(cudaResourceDesc));texRes.resType cudaResourceTypeArray;texRes.res.array.array m_pRFData;cudaTextureDesc texDescr;memset(texDescr, 0, sizeof(cudaTextureDesc));texDescr.normalizedCoords false;texDescr.filterMode cudaFilterModeLinear; // 这里很重要texDescr.addressMode[0] cudaAddressModeBorder;texDescr.addressMode[1] cudaAddressModeBorder;error cudaCreateTextureObject(m_tex, texRes, texDescr, NULL);//int16_t pi16Src[nRx * Nsample * IQ] {1, 11, 2, 22,// 3, 33, 4, 44, // 5, 55, 6, 66, // 7, 77, 8, 88};//int16_t pi16Src[nRx * Nsample * IQ] { 1, 11, 2, 22,// 3, 33, 4, 44};int16_t pi16Src[nRx * Nsample * IQ] { 1,2,3,4 };error cudaMalloc(m_i16RFDataBuffer, sizeof(int16_t) * nRx * IQ * Nsample);error cudaMemcpy(m_i16RFDataBuffer, pi16Src, sizeof(int16_t) * nRx * IQ * Nsample, cudaMemcpyHostToDevice);error cudaMalloc(m_pHalfRFDataCache, sizeof(half) * nRx * IQ * Nsample);Data2Half(m_pHalfRFDataCache, m_i16RFDataBuffer, nRx * IQ * Nsample);error cudaMemcpy2DToArray(m_pRFData, 0, 0, m_pHalfRFDataCache, sizeof(half) * nRx * IQ, sizeof(half) * nRx * IQ, Nsample, cudaMemcpyDeviceToDevice);float* pf_res1 nullptr;float* pf_res2 nullptr;error cudaMalloc(pf_res1, nRx * Nsample * sizeof(float)); cudaMemset(pf_res1, 0, nRx * Nsample * sizeof(float));error cudaMalloc(pf_res2, nRx * Nsample * sizeof(float)); cudaMemset(pf_res2, 0, nRx * Nsample * sizeof(float));error cudaGetLastError();dim3 block_dim dim3(1, 1);dim3 grid_dim dim3(1, 1);Tex2DTest grid_dim, block_dim (m_tex, pf_res1, pf_res2);cudaDeviceSynchronize();std::vectorfloat vf_res_1(nRx * Nsample, 0);std::vectorfloat vf_res_2(nRx * Nsample, 0);cudaMemcpy(vf_res_1.data(), pf_res1, sizeof(float) * vf_res_1.size(), cudaMemcpyDeviceToHost);cudaMemcpy(vf_res_2.data(), pf_res2, sizeof(float) * vf_res_2.size(), cudaMemcpyDeviceToHost);return 0; }void Data2Half(half* pDst, const int16_t* pSrc, const int Ndots) {dim3 block dim3(512, 1);dim3 grid dim3((Ndots - 1) / block.x 1, 1);data2half grid, block (pDst, pSrc, Ndots); }static __global__ void Tex2DTest(cudaTextureObject_t p_rf_data, float *pfRes1, float *pfRes2) {for (size_t y 0; y 2; y){for (size_t x 0; x 2; x) {float value tex2Dfloat(p_rf_data, x, y);//pfRes1[y * 4 y] printf(x: %f\n, value);}} }3. 输出分析 可以看到执行结果是 为什么呢 原因是因为tex2D插值导致的上面测试数据是 1  2 3   4 那在进行插值的时候会变成 0  0   0   0 0   1   2  0 0   3   4  0 每个点的输出都是当前前和左上角3个点进行平均计算出来的 比如第一个输出计算为1 0 0 0/4 0.25 最后一个输出的计算为1 2 3 4 / 4 2.5 4. 问题 上面只是单独数据实数点的计算如果我的数据集合是复数怎么办 比如一组2 * 2大小的数据对 1 2 3 4; 5,   6 7 8 数据实际表示含义是 1 j * 2,   3 j * 4; 5 j * 6,   7 j * 8 这种情况下怎么做到正确插值呢比如第一个实数点的输出结果应该是 1 0 0 0/ 4 最后一个实数点的输出应该是 1 3 5 7 / 4 同理最后一个虚数点的输出应该是            2 4 6 8/ 4 5. 解决 #include cuda_runtime.h #include device_launch_parameters.h #include assert.h #include stdio.h #include iostream #include cuda_fp16.h #include vectorvoid Data2Half(half* pDst, const int16_t* pSrc, const int Ndots); static __global__ void Tex2DTest(cudaTextureObject_t p_rf_data, float* pfRes1, float* pfRes2);static __global__ void data2half(half* pDst, const int16_t* pSrc, const int Ndots) {const int tid blockIdx.x * blockDim.x threadIdx.x;if (tid Ndots)return;pDst[tid] __short2half_rn(pSrc[tid]); }cudaTextureObject_t m_tex 0; cudaArray* m_pRFData nullptr; int16_t* m_i16RFDataBuffer nullptr; // 设备端的RF数据 half* m_pHalfRFDataCache nullptr; // 转换为半浮点型的RF数据缓存用于将SHORT类型转换为FLOAT类型using namespace std;int main() {const int nRx 2;const int Nsample 2;const int IQ 2;cudaError_t error;cudaChannelFormatDesc channelDesc cudaCreateChannelDescHalf2();error cudaMallocArray(m_pRFData, channelDesc, nRx, Nsample, cudaArrayTextureGather);assert(m_pRFData);cudaResourceDesc texRes;memset(texRes, 0, sizeof(cudaResourceDesc));texRes.resType cudaResourceTypeArray;texRes.res.array.array m_pRFData;cudaTextureDesc texDescr;memset(texDescr, 0, sizeof(cudaTextureDesc));texDescr.normalizedCoords false;texDescr.filterMode cudaFilterModeLinear; // 这里很重要texDescr.addressMode[0] cudaAddressModeBorder;texDescr.addressMode[1] cudaAddressModeBorder;error cudaCreateTextureObject(m_tex, texRes, texDescr, NULL);//int16_t pi16Src[nRx * Nsample * IQ] {1, 11, 2, 22,// 3, 33, 4, 44, // 5, 55, 6, 66, // 7, 77, 8, 88};//int16_t pi16Src[nRx * Nsample * IQ] { 1, 11, 2, 22,// 3, 33, 4, 44};int16_t pi16Src[nRx * Nsample * IQ] { 1, 2, 3, 4,5, 6, 7, 8 };error cudaMalloc(m_i16RFDataBuffer, sizeof(int16_t) * nRx * IQ * Nsample);error cudaMemcpy(m_i16RFDataBuffer, pi16Src, sizeof(int16_t) * nRx * IQ * Nsample, cudaMemcpyHostToDevice);error cudaMalloc(m_pHalfRFDataCache, sizeof(half) * nRx * IQ * Nsample);Data2Half(m_pHalfRFDataCache, m_i16RFDataBuffer, nRx * IQ * Nsample);error cudaMemcpy2DToArray(m_pRFData, 0, 0, m_pHalfRFDataCache, sizeof(half2) * nRx, sizeof(half2) * nRx, Nsample, cudaMemcpyDeviceToDevice);float* pf_res1 nullptr;float* pf_res2 nullptr;error cudaMalloc(pf_res1, nRx * Nsample * sizeof(float)); cudaMemset(pf_res1, 0, nRx * Nsample * sizeof(float));error cudaMalloc(pf_res2, nRx * Nsample * sizeof(float)); cudaMemset(pf_res2, 0, nRx * Nsample * sizeof(float));error cudaGetLastError();dim3 block_dim dim3(1, 1);dim3 grid_dim dim3(1, 1);Tex2DTest grid_dim, block_dim (m_tex, pf_res1, pf_res2);cudaDeviceSynchronize();std::vectorfloat vf_res_1(nRx * Nsample, 0);std::vectorfloat vf_res_2(nRx * Nsample, 0);cudaMemcpy(vf_res_1.data(), pf_res1, sizeof(float) * vf_res_1.size(), cudaMemcpyDeviceToHost);cudaMemcpy(vf_res_2.data(), pf_res2, sizeof(float) * vf_res_2.size(), cudaMemcpyDeviceToHost);return 0; }void Data2Half(half* pDst, const int16_t* pSrc, const int Ndots) {dim3 block dim3(512, 1);dim3 grid dim3((Ndots - 1) / block.x 1, 1);data2half grid, block (pDst, pSrc, Ndots); }static __global__ void Tex2DTest(cudaTextureObject_t p_rf_data, float* pfRes1, float* pfRes2) {for (size_t y 0; y 2; y){for (size_t x 0; x 2; x){float2 value tex2Dfloat2(p_rf_data, x, y);//pfRes1[y * 4 y] printf(x: %f, y: %f, value.x, value.y);// printf(x: %f, y: %f\n, value.x, value.y);}printf(\n);} }其实关键是在tex2D的构造 然后按照half2的方式进行排布就好了