最近调用FFTW来做reconstruction，遇到一个问题，先贴code：
代码
2 fftw_complex in[N], out[N];
3 fftw_plan p;
4
5 p = fftw_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_ESTIMATE);
6
7 // 初始化in
8 for (int i = 0; i < N; i++)
9 {
10 in[i][0] = i; // 实数部分
11 in[i][1] = 0; // 虚数部分
12 }
13
14
15 // 执行fftw
16 fftw_execute(p);
17
18 fftw_complex out1[N];
19
20 fftw_plan p1;
21
22 p1 = fftw_plan_dft_1d(N, out, out1, FFTW_BACKWARD, FFTW_ESTIMATE);
23
24 fftw_execute(p1);
25
26 fftw_destroy_plan(p);
27 fftw_destroy_plan(p1);
28
29 double residual = 0;
30
31 for(int i = 0; i < N; i++)
32 {
33 out1[i][0] /= N;
34 out1[i][1] /= N;
35
36 residual += (out1[i][0] - in[i][0])*(out1[i][0] - in[i][0])
37 + (out1[i][1] - in[i][1])*(out1[i][1] - in[i][1]);
38 }
39
40
41 if (residual < 1e-6)
42 cout << "1D FFT passed successful!" << " --- " << "residual : " << residual << endl;
43 else
44 cout << "1D FFT failed!" << " --- " << "residual : " << residual << endl;
可是，却发现，如果如下调用
p = fftw_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_MEASURE);
则结果会以一定的百分比运行成功，大概是运行十次，成功5次，成功几率50%.
而如果以
p = fftw_plan_dft_1d(N, in, out, FFTW_FORWARD, FFTW_ESTIMATE);
则，结果总是运行成功。
查了一下，官方解释如下：
The flags argument is usually either FFTW_MEASURE or FFTW_ESTIMATE. FFTW_MEASURE instructs FFTW to run and measure the execution time of several FFTs in order to find the best way to compute the transform of size n. This process takes some time (usually a few seconds), depending on your machine and on the size of the transform. FFTW_ESTIMATE, on the contrary, does not run any computation and just builds a reasonable plan that is probably sub-optimal. In short, if your program performs many transforms of the same size and initialization time is not important, use FFTW_MEASURE; otherwise use the estimate. The data in the in/out arrays is overwritten during FFTW_MEASURE planning, so such planning should be done before the input is initialized by the user.
原因还需要进一步找。感觉还是有问题滴。