I'm trying to write an R wrapper for the FINUFFT routines for calculating the FFT of an unevenly sampled series. I have virtually no experience with C/C++, so I'm working from an example that compares the traditional Fourier transform to the NUFFT. The example code follows.
// this is all you must include for the finufft lib...
#include "finufft.h"
#include <complex>
// also needed for this example...
#include <stdio.h>
#include <stdlib.h>
using namespace std;
int main(int argc, char* argv[])
/* Simple example of calling the FINUFFT library from C++, using plain
arrays of C++ complex numbers, with a math test. Barnett 3/10/17
Double-precision version (see example1d1f for single-precision)
Compile with:
g++ -fopenmp example1d1.cpp -I ../src ../lib-static/libfinufft.a -o example1d1 -lfftw3 -lfftw3_omp -lm
or if you have built a single-core version:
g++ example1d1.cpp -I ../src ../lib-static/libfinufft.a -o example1d1 -lfftw3 -lm
Usage: ./example1d1
*/
{
int M = 1e6; // number of nonuniform points
int N = 1e6; // number of modes
double acc = 1e-9; // desired accuracy
nufft_opts opts; finufft_default_opts(&opts);
complex<double> I = complex<double>(0.0,1.0); // the imaginary unit
// generate some random nonuniform points (x) and complex strengths (c):
double *x = (double *)malloc(sizeof(double)*M);
complex<double>* c = (complex<double>*)malloc(sizeof(complex<double>)*M);
for (int j=0; j<M; ++j) {
x[j] = M_PI*(2*((double)rand()/RAND_MAX)-1); // uniform random in [-pi,pi)
c[j] = 2*((double)rand()/RAND_MAX)-1 + I*(2*((double)rand()/RAND_MAX)-1);
}
// allocate output array for the Fourier modes:
complex<double>* F = (complex<double>*)malloc(sizeof(complex<double>)*N);
// call the NUFFT (with iflag=+1): note N and M are typecast to BIGINT
int ier = finufft1d1(M,x,c,+1,acc,N,F,opts);
int n = 142519; // check the answer just for this mode...
complex<double> Ftest = complex<double>(0,0);
for (int j=0; j<M; ++j)
Ftest += c[j] * exp(I*(double)n*x[j]);
int nout = n+N/2; // index in output array for freq mode n
double Fmax = 0.0; // compute inf norm of F
for (int m=0; m<N; ++m) {
double aF = abs(F[m]);
if (aF>Fmax) Fmax=aF;
}
double err = abs(F[nout] - Ftest)/Fmax;
printf("1D type-1 NUFFT done. ier=%d, err in F[%d] rel to max(F) is %.3g\n",ier,n,err);
free(x); free(c); free(F);
return ier;
}
Much of this I don't need, such as generating the test series and comparing to the traditional FFT. Further, I want to return the values of the transform, not just an error code indicating success. Below is my code.
#include "finufft.h"
#include <complex>
#include <Rcpp.h>
#include <stdlib.h>
using namespace Rcpp;
using namespace std;
// [[Rcpp::export]]
ComplexVector finufft(int M, NumericVector x, ComplexVector c, int N) {
// From example code for finufft, sets precision and default options
double acc = 1e-9;
nufft_opts opts; finufft_default_opts(&opts);
// allocate output array for the finufft routine:
complex<double>* F = (complex<double>*)malloc(sizeof(complex<double>*)*N);
// Change vector inputs from R types to C++ types
double* xd = as< double* >(x);
complex<double>* cd = as< complex<double>* >(c);
// call the NUFFT (with iflag=-1): note N and M are typecast to BIGINT
int ier = finufft1d1(M,xd,cd,-1,acc,N,F,opts);
ComplexVector Fd = as<ComplexVector>(*F);
return Fd;
}
When I try to source this in Rstudio, I get the error "no matching function for call to 'as(std::complex<double>*&)'", pointing to the line declaring Fd towards the end. I believe the error indicates that either the function 'as' isn't defined (which I know is false), or the argument to 'as' isn't the correct type. The examples here include one using 'as' to convert to a NumericVector, so unless there's some complication with complex values I don't see why it should be a problem here.
I know there are potential problems using two namespaces, but I don't believe that's the issue here. My best guess is that there's an issue with how I'm trying to use pointers, but I lack the experience to identify it and I can't find any similar examples online to guide me.
Rcpp::as<T>从R数据类型（SEXP）转换为C ++数据类型，例如Rcpp::ComplexVector。这不适合您尝试从C样式数组转换为C ++的情况。幸运的是Rcpp::Vector，它的基础Rcpp::ComplexVector是此任务的构造函数：Vector (InputIterator first, InputIterator last)。对于另一个方向（从C ++到C样式数组），可以使用vector.begin()或&vector[0]。
但是，需要reinterpret_cast在Rcomplex*和之间进行转换std::complex<double>*。但是，这应该不会造成任何问题，因为Rcomplex（aka complex double在C中）并且std::complex<doulbe>是兼容的。
一个最小的例子：
#include <Rcpp.h>
#include <complex>
using namespace Rcpp;
// [[Rcpp::export]]
ComplexVector foo(ComplexVector v) {
std::complex<double>* F = reinterpret_cast<std::complex<double>*>(v.begin());
int N = v.length();
// do something with F
ComplexVector Fd(reinterpret_cast<Rcomplex*>(F),
reinterpret_cast<Rcomplex*>(F + N));
return Fd;
}
/*** R
set.seed(42)
foo(runif(4)*(1+1i))
*/
结果：
> Rcpp::sourceCpp('56675308/code.cpp')
> set.seed(42)
> foo(runif(4)*(1+1i))
[1] 0.9148060+0.9148060i 0.9370754+0.9370754i 0.2861395+0.2861395i 0.8304476+0.8304476i
顺便说一句，您可以将它们用作函数的参数和返回类型，从而将它们reinterpret_cast移出视线std::vector<std::complex<double>>。Rcpp为您完成其余的工作。这也有助于摆脱裸体malloc：
#include <Rcpp.h>
// dummy function with reduced signature
int finufft1d1(int M, double *xd, std::complex<double> *cd, int N, std::complex<double> *Fd) {
return 0;
}
// [[Rcpp::export]]
std::vector<std::complex<double>> finufft(int M,
std::vector<double> x,
std::vector<std::complex<double>> c,
int N) {
// allocate output array for the finufft routine:
std::vector<std::complex<double>> F(N);
// Change vector inputs from R types to C++ types
double* xd = x.data();
std::complex<double>* cd = c.data();
std::complex<double>* Fd = F.data();
int ier = finufft1d1(M, xd, cd, N, Fd);
return F;
}