错误的结果cufft 3D就地

本文介绍了错误的结果cufft 3D就地的处理方法，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！ 我写，因为我面对的问题与袖口3D转换就地，而我没有问题的外部版本。我试图关注Robert Crovella的回答这里，但是我没有获得正确的结果，当我做FFT + IFT。 这是我的代码： #include< stdio.h> #include< stdlib.h> #include< cuda_runtime.h> #include< complex.h> #include< cuComplex.h> #include< cufft.h> //主函数 int main（int argc，char ** argv）{ int N = 4; double * in = NULL，* d_in = NULL; cuDoubleComplex * out = NULL，* d_out = NULL; cufftHandle plan_r2c，plan_c2r; unsigned int out_mem_size = sizeof（cuDoubleComplex）* N * N *（N / 2 + 1）; unsigned int in_mem_size = out_mem_size; in =（double *）malloc（in_mem_size）; out =（cuDoubleComplex *）in; cudaMalloc（（void **）& d_in，in_mem_size）; d_out =（cuDoubleComplex *）d_in; cufftPlan3d（& plan_r2c，N，N，N，CUFFT_D2Z）; cufftPlan3d（& plan_c2r，N，N，N，CUFFT_Z2D）; memset（in，0，in_mem_size）; unsigned int idx; for（int z = 0; z for（int y = 0; y for（int x = 0; x < N; x ++）{ idx = z + N *（y + x * N）; in [idx] = idx; } } } printf（\\\Start：\\\）; for（int z = 0; z< N; z ++）{ printf（plane =％d -------------------- -------- \\\，z）; for（int x = 0; x for（int y = 0; y idx = z + N * + x * N）; printf（％。3f \t，in [idx]）; } printf（\\\）; } } cudaMemcpy（d_in，in，in_mem_size，cudaMemcpyHostToDevice）; cufftExecD2Z（plan_r2c，（cufftDoubleReal *）d_in，（cufftDoubleComplex *）d_out）; cufftExecZ2D（plan_c2r，（cufftDoubleComplex *）d_out，（cufftDoubleReal *）d_in）; memset（in，0，in_mem_size）; CU_ERR_CHECK（cudaMemcpy（in，d_in，in_mem_size，cudaMemcpyDeviceToHost））; printf（\\\After FFT + IFT：\\\）; for（int z = 0; z< N; z ++）{ printf（plane =％d -------------------- -------- \\\，z）; for（int x = 0; x for（int y = 0; y idx = z + N * + x * N）; //规范化 in [idx] / =（N * N * N）; printf（％。3f \t，in [idx]）; } printf（\\\）; } } return 0; } 程序输出以下数据： 启动文件 plane = 0 ------------------------ ---- 0.000 4.000 8.000 12.000 16.000 20.000 24.000 28.000 32.000 36.000 40.000 44.000 48.000 52.000 56.000 60.000 plane = 1 -------------------------- - 1.000 5.000 9.000 13.000 17.000 21.000 25.000 29.000 33.000 37.000 41.000 45.000 49.000 53.000 57.000 61.000 plane = 2 ---------------------------- 2.000 6.000 10.000 14.000 18.000 22.000 26.000 30.000 34.000 38.000 42.000 46.000 50.000 54.000 58.000 62.000 plane = 3 ---------------------------- p> 3.000 7.000 11.000 15.000 19.000 23.000 27.000 31.000 35.000 39.000 43.000 47.000 51.000 55.000 59.000 63.000 FFT + IFT后 plane = 0 ------------- --------------- -0.000 -0.344 8.000 12.000 -0.031 20.000 24.000 -0.031 32.000 36.000 0.031 44.000 48.000 -0.094 56.000 60.000 plane = 1 ---------- ------------------ 1.000 -0.000 9.000 13.000 -0.000 21.000 25.000 0.125 33.000 37.000 0.000 45.000 49.000 0.000 57.000 61.000 plane = 2 ---------- ------------------ 2.000 6.000 -0.000 14.000 18.000 0.000 26.000 30.000 0.000 38.000 42.000 -0.000 50.000 54.000 -0.000 62.000 plane = 3 --------- ------------------- 3.000 7.000 0.031 15.000 19.000 -0.031 27.000 31.000 -0.031 39.000 43.000 0.031 51.000 55.000 0.031 63.000 我甚至尝试以这种方式填充数据： //带有填充 unsigned int idx; for（int x = 0; x for（int y = 0; y for（int z = 0; z < 2 *（N / 2 + 1）; z ++）{ idx = z + N *（y + x * N） if（z else in [idx] = 0; } } } 我做错了什么？ 解决方案正如您已经发现的，如果使用 CUFFT_COMPATIBILITY_FFTW_PADDING 默认的兼容模式。要使代码生效，您可以使用 cufftSetCompatibilityMode（）设置 CUFFT_COMPATIBILITY_NATIVE 。但是，此模式在当前版本的CUDA中标记为已弃用。 因此，我建议使用默认兼容模式并使用填充。您尝试实现填充是错误的。计算3维x，y，z的线性指数的公式其中z是最快运行指数， idx = z + Nz *（y + Ny * x）。包括填充的 z 维的大小 Nz 是 Nz =（N / 2 + 1 ）* 2 。然后，数组的正确初始化是： unsigned int idx; for（int z = 0; z for（int y = 0; y for（int x = 0; x < N; x ++）{ idx = z +（N / 2 + 1）* 2 *（y + x * N） in [idx] = idx; } } } I write because I'm facing problems with the cufft 3D transform in-place, while I have no problems for the out-of-place version. I tried to follow Robert Crovella's answer here but I'm not obtaining the correct results when I make a FFT+IFT.This is my code:#include <stdio.h>#include <stdlib.h>#include <cuda_runtime.h>#include <complex.h>#include <cuComplex.h>#include <cufft.h>// Main functionint main(int argc, char **argv){ int N = 4; double *in = NULL, *d_in = NULL; cuDoubleComplex *out = NULL, *d_out = NULL; cufftHandle plan_r2c, plan_c2r; unsigned int out_mem_size = sizeof(cuDoubleComplex) * N*N*(N/2 + 1); unsigned int in_mem_size = out_mem_size; in = (double *) malloc (in_mem_size); out = (cuDoubleComplex *)in; cudaMalloc((void **)&d_in, in_mem_size); d_out = (cuDoubleComplex *)d_in; cufftPlan3d(&plan_r2c, N, N, N, CUFFT_D2Z); cufftPlan3d(&plan_c2r, N, N, N, CUFFT_Z2D); memset(in, 0, in_mem_size); unsigned int idx; for (int z = 0; z < N; z++){ for (int y = 0; y < N; y++){ for (int x = 0; x < N; x++){ idx = z + N * ( y + x * N); in[idx] = idx; } } } printf("\nStart: \n"); for (int z = 0; z < N; z++){ printf("plane = %d ----------------------------\n", z); for (int x = 0; x < N; x++){ for (int y = 0; y < N; y++){ idx = z + N * ( y + x * N); printf("%.3f \t", in[idx]); } printf("\n"); } } cudaMemcpy(d_in, in, in_mem_size, cudaMemcpyHostToDevice); cufftExecD2Z(plan_r2c, (cufftDoubleReal *)d_in, (cufftDoubleComplex *)d_out); cufftExecZ2D(plan_c2r, (cufftDoubleComplex *)d_out, (cufftDoubleReal *)d_in); memset(in, 0, in_mem_size); CU_ERR_CHECK( cudaMemcpy(in, d_in, in_mem_size, cudaMemcpyDeviceToHost) ); printf("\nAfter FFT+IFT: \n"); for (int z = 0; z < N; z++){ printf("plane = %d ----------------------------\n", z); for (int x = 0; x < N; x++){ for (int y = 0; y < N; y++){ idx = z + N * ( y + x * N); // Normalisation in[idx] /= (N*N*N); printf("%.3f \t", in[idx]); } printf("\n"); } } return 0;}The program outputs the following data:Starting fileplane = 0 ----------------------------0.000 4.000 8.000 12.00016.000 20.000 24.000 28.00032.000 36.000 40.000 44.00048.000 52.000 56.000 60.000plane = 1 ----------------------------1.000 5.000 9.000 13.00017.000 21.000 25.000 29.00033.000 37.000 41.000 45.00049.000 53.000 57.000 61.000plane = 2 ----------------------------2.000 6.000 10.000 14.00018.000 22.000 26.000 30.00034.000 38.000 42.000 46.00050.000 54.000 58.000 62.000plane = 3 ----------------------------3.000 7.000 11.000 15.00019.000 23.000 27.000 31.00035.000 39.000 43.000 47.00051.000 55.000 59.000 63.000After FFT+IFTplane = 0 -----------------------------0.000 -0.344 8.000 12.000-0.031 20.000 24.000 -0.03132.000 36.000 0.031 44.00048.000 -0.094 56.000 60.000plane = 1 ----------------------------1.000 -0.000 9.000 13.000-0.000 21.000 25.000 0.12533.000 37.000 0.000 45.00049.000 0.000 57.000 61.000plane = 2 ----------------------------2.000 6.000 -0.000 14.00018.000 0.000 26.000 30.0000.000 38.000 42.000 -0.00050.000 54.000 -0.000 62.000plane = 3 ----------------------------3.000 7.000 0.031 15.00019.000 -0.031 27.000 31.000-0.031 39.000 43.000 0.03151.000 55.000 0.031 63.000I even tried to pad the data this way:// With padding unsigned int idx; for (int x = 0; x < N; x++){ for (int y = 0; y < N; y++){ for (int z = 0; z < 2*(N/2+1); z++){ idx = z + N * ( y + x * N); if (z < 4) in[idx] = idx; else in[idx] = 0; } } }What am I doing wrong? 解决方案 As you already found out, you need padding if you use the CUFFT_COMPATIBILITY_FFTW_PADDINGcompatibility mode which is default. For your code to work you could use cufftSetCompatibilityMode() to set CUFFT_COMPATIBILITY_NATIVE. However, this mode is marked as deprecated in the current version of CUDA.Therefore, I recommend to use the default compatibility mode and use padding. Your try to implement padding is wrong. The formula to calculate a linear index for 3 dimension x, y, z where z is the fastest running index is idx = z + Nz*(y + Ny*x). The size Nz of the z dimension including padding is Nz = (N/2+1)*2. Then, the correct initialization of the array is:unsigned int idx;for (int z = 0; z < N; z++){ for (int y = 0; y < N; y++){ for (int x = 0; x < N; x++){ idx = z + (N/2+1)*2 * ( y + x * N); in[idx] = idx; } }}Accordingly for the print loops. 这篇关于错误的结果cufft 3D就地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！