Factored discrete Fourier transform, or FFT, and its inverse iFFT - http://www.math.wustl.edu/~victor/mfmm/fourier/

fftifft.c

#include <assert.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>

#define q	3		/* for 2^3 points */
#define N	(1<<q)		/* N-point FFT, iFFT */

typedef float real;
typedef struct{
  real Re;
  real Im;
} complex;

#ifndef PI
# define PI	3.14159265358979323846264338327950288
#endif

/* Print a vector of complexes as ordered pairs. */
static void print_vector(const char *title, complex *x, int n)
{
  int i;
  printf("%s (dim=%d):", title, n);
  for(i=0; i<n; i++ )
    printf(" %5.2f,%5.2f ", x[i].Re,x[i].Im);
  putchar('\n');
  return;
}

/*
   fft(v,N):
   [0] If N==1 then return.
   [1] For k = 0 to N/2-1, let ve[k] = v[2*k]
   [2] Compute fft(ve, N/2);
   [3] For k = 0 to N/2-1, let vo[k] = v[2*k+1]
   [4] Compute fft(vo, N/2);
   [5] For m = 0 to N/2-1, do [6] through [9]
   [6]   Let w.re = cos(2*PI*m/N)
   [7]   Let w.im = -sin(2*PI*m/N)
   [8]   Let v[m] = ve[m] + w*vo[m]
   [9]   Let v[m+N/2] = ve[m] - w*vo[m]
 */
void fft( complex *v, int n, complex *tmp )
{
  if(n>1) {			/* otherwise, do nothing and return */
    int k,m;    complex z, w, *vo, *ve;
    ve = tmp; vo = tmp+n/2;
    for(k=0; k<n/2; k++) {
      ve[k] = v[2*k];
      vo[k] = v[2*k+1];
    }
    fft( ve, n/2, v );		/* FFT on even-indexed elements of v[] */
    fft( vo, n/2, v );		/* FFT on odd-indexed elements of v[] */
    for(m=0; m<n/2; m++) {
      w.Re = cos(2*PI*m/(double)n);
      w.Im = -sin(2*PI*m/(double)n);
      z.Re = w.Re*vo[m].Re - w.Im*vo[m].Im;	/* Re(w*vo[m]) */
      z.Im = w.Re*vo[m].Im + w.Im*vo[m].Re;	/* Im(w*vo[m]) */
      v[  m  ].Re = ve[m].Re + z.Re;
      v[  m  ].Im = ve[m].Im + z.Im;
      v[m+n/2].Re = ve[m].Re - z.Re;
      v[m+n/2].Im = ve[m].Im - z.Im;
    }
  }
  return;
}

/*
   ifft(v,N):
   [0] If N==1 then return.
   [1] For k = 0 to N/2-1, let ve[k] = v[2*k]
   [2] Compute ifft(ve, N/2);
   [3] For k = 0 to N/2-1, let vo[k] = v[2*k+1]
   [4] Compute ifft(vo, N/2);
   [5] For m = 0 to N/2-1, do [6] through [9]
   [6]   Let w.re = cos(2*PI*m/N)
   [7]   Let w.im = sin(2*PI*m/N)
   [8]   Let v[m] = ve[m] + w*vo[m]
   [9]   Let v[m+N/2] = ve[m] - w*vo[m]
 */
void ifft( complex *v, int n, complex *tmp )
{
  if(n>1) {			/* otherwise, do nothing and return */
    int k,m;    complex z, w, *vo, *ve;
    ve = tmp; vo = tmp+n/2;
    for(k=0; k<n/2; k++) {
      ve[k] = v[2*k];
      vo[k] = v[2*k+1];
    }
    ifft( ve, n/2, v );		/* FFT on even-indexed elements of v[] */
    ifft( vo, n/2, v );		/* FFT on odd-indexed elements of v[] */
    for(m=0; m<n/2; m++) {
      w.Re = cos(2*PI*m/(double)n);
      w.Im = sin(2*PI*m/(double)n);
      z.Re = w.Re*vo[m].Re - w.Im*vo[m].Im;	/* Re(w*vo[m]) */
      z.Im = w.Re*vo[m].Im + w.Im*vo[m].Re;	/* Im(w*vo[m]) */
      v[  m  ].Re = ve[m].Re + z.Re;
      v[  m  ].Im = ve[m].Im + z.Im;
      v[m+n/2].Re = ve[m].Re - z.Re;
      v[m+n/2].Im = ve[m].Im - z.Im;
    }
  }
  return;
}

int main(void)
{
  complex v[N], v1[N], scratch[N];
  int k;

  /* Fill v[] with a function of known FFT: */
  for(k=0; k<N; k++) {
    v[k].Re = 0.125*cos(2*PI*k/(double)N);
    v[k].Im = 0.125*sin(2*PI*k/(double)N);
    v1[k].Re =  0.3*cos(2*PI*k/(double)N);
    v1[k].Im = -0.3*sin(2*PI*k/(double)N);
  }

  /* FFT, iFFT of v[]: */
  print_vector("Orig", v, N);
  fft( v, N, scratch );
  print_vector(" FFT", v, N);
  ifft( v, N, scratch );
  print_vector("iFFT", v, N);

  /* FFT, iFFT of v1[]: */
  print_vector("Orig", v1, N);
  fft( v1, N, scratch );
  print_vector(" FFT", v1, N);
  ifft( v1, N, scratch );
  print_vector("iFFT", v1, N);

  exit(EXIT_SUCCESS);
}