int WebRtcIsac_Fftns(unsigned int ndim, const int dims[],
double Re[],
double Im[],
int iSign,
double scaling,
FFTstr *fftstate)
{
size_t nSpan, nPass, nTotal;
unsigned int i;
int ret, max_factors, max_perm;
/*
* tally the number of elements in the data array
* and determine the number of dimensions
*/
nTotal = 1;
if (ndim && dims [0])
{
for (i = 0; i < ndim; i++)
{
if (dims [i] <= 0)
{
return -1;
}
nTotal *= dims [i];
}
}
else
{
ndim = 0;
for (i = 0; dims [i]; i++)
{
if (dims [i] <= 0)
{
return -1;
}
nTotal *= dims [i];
ndim++;
}
}
/* determine maximum number of factors and permuations */
#if 1
/*
* follow John Beale's example, just use the largest dimension and don't
* worry about excess allocation. May be someone else will do it?
*/
max_factors = max_perm = 1;
for (i = 0; i < ndim; i++)
{
nSpan = dims [i];
if ((int)nSpan > max_factors)
{
max_factors = (int)nSpan;
}
if ((int)nSpan > max_perm)
{
max_perm = (int)nSpan;
}
}
#else
/* use the constants used in the original Fortran code */
max_factors = 23;
max_perm = 209;
#endif
/* loop over the dimensions: */
nPass = 1;
for (i = 0; i < ndim; i++)
{
nSpan = dims [i];
nPass *= nSpan;
ret = FFTRADIX (Re, Im, nTotal, nSpan, nPass, iSign,
max_factors, max_perm, fftstate);
/* exit, clean-up already done */
if (ret)
return ret;
}
/* Divide through by the normalizing constant: */
if (scaling && scaling != 1.0)
{
if (iSign < 0) iSign = -iSign;
if (scaling < 0.0)
{
scaling = (double)nTotal;
if (scaling < -1.0)
scaling = sqrt (scaling);
}
scaling = 1.0 / scaling; /* multiply is often faster */
for (i = 0; i < nTotal; i += iSign)
{
Re [i] *= scaling;
Im [i] *= scaling;
}
}
return 0;
}
int
FFTN (int ndim,
const unsigned int dims [],
REAL Re [],
REAL Im [],
int iSign,
double scaling)
{
unsigned int nTotal;
unsigned int maxFactors, maxPerm;
/*
* tally the number of elements in the data array
* and determine the number of dimensions
*/
nTotal = 1;
if (ndim)
{
if (dims != NULL)
{
int i;
/* number of dimensions was specified */
for (i = 0; i < ndim; i++)
{
if (dims [i] <= 0) goto Dimension_Error;
nTotal *= dims [i];
}
}
else
nTotal *= ndim;
}
else
{
int i;
/* determine # of dimensions from zero-terminated list */
if (dims == NULL) goto Dimension_Error;
for (ndim = i = 0; dims [i]; i++)
{
if (dims [i] <= 0)
goto Dimension_Error;
nTotal *= dims [i];
ndim++;
}
}
/* determine maximum number of factors and permuations */
#if 1
/*
* follow John Beale's example, just use the largest dimension and don't
* worry about excess allocation. May be someone else will do it?
*/
if (dims != NULL)
{
int i;
for (maxFactors = maxPerm = 1, i = 0; i < ndim; i++)
{
if (dims [i] > maxFactors) maxFactors = dims [i];
if (dims [i] > maxPerm) maxPerm = dims [i];
}
}
else
{
maxFactors = maxPerm = nTotal;
}
#else
/* use the constants used in the original Fortran code */
maxFactors = 23;
maxPerm = 209;
#endif
/* loop over the dimensions: */
if (dims != NULL)
{
unsigned int nSpan = 1;
int i;
for (i = 0; i < ndim; i++)
{
int ret;
nSpan *= dims [i];
ret = FFTRADIX (Re, Im, nTotal, dims [i], nSpan, iSign,
maxFactors, maxPerm);
/* exit, clean-up already done */
if (ret)
return ret;
}
}
else
{
int ret;
ret = FFTRADIX (Re, Im, nTotal, nTotal, nTotal, iSign,
maxFactors, maxPerm);
/* exit, clean-up already done */
if (ret)
return ret;
}
/* Divide through by the normalizing constant: */
if (scaling && scaling != 1.0)
{
size_t ist;
if (iSign < 0) iSign = -iSign;
if (scaling < 0.0)
scaling = (scaling < -1.0) ? sqrt (nTotal) : nTotal;
scaling = 1.0 / scaling; /* multiply is often faster */
for (ist = 0; ist < nTotal; ist += iSign)
{
Re_Data (ist) *= scaling;
Im_Data (ist) *= scaling;
}
}
return 0;
Dimension_Error:
fprintf (stderr, "Error: " FFTNS "() - dimension error\n");
fft_free (); /* free-up memory */
return -1;
}
