Uhh, of course. I forgot that (5>>1)<<1 != 5 ...
Thanks for the clarification!
Best,
Alex.
On Thursday, 19 June 2014 12:16:29 UTC+2, Tim Holy wrote:
>
> On Thursday, June 19, 2014 02:59:45 AM Alex wrote:
> > Looking at the code of brfft (in fftw.jl) I find
> >
> > function brfft(X::StridedArray{$Tc}, d::Integer, region::Integer)
> > osize = [size(X)...]
> > @assert osize[region] == d>>1 + 1
> > osize[region] = d
> > # snip ...
> >
> > which seems to imply that d can always be calculated from
> size(X,region)?
>
> You're checking that the value of d is consistent, you're not calculating
> it.
> Say osize[region] == 3. d could be either 4 or 5.
>
> --Tim
>
> >
> > - Alex.
> >
> > On Thursday, 19 June 2014 11:32:59 UTC+2, Tim Holy wrote:
> > > On Thursday, June 19, 2014 12:27:40 AM Alex wrote:
> > > > Actually, it is not really clear to me why `d` AKA `len` is needed
> at
> > >
> > > all,
> > >
> > > > since d = (size(A,dim[1])-1)<<1 appears to be well defined. I am
> > >
> > > probably
> > >
> > > > missing something here. Maybe Steven or someone else can clarify
> this?
> > >
> > > Doc bug. The A referred to in this statement is the _original
> real-valued_
> > > A.
> > > And the formula is wrong. A better way to write the help would be
> > >
> > > Base.irfft(Afft, d[, dims])
> > >
> > > Inverse of "rfft()": for a complex array "Afft", gives the
> > > corresponding real array "A" whose FFT yields "Afft" in the first
> half.
> > > As for "rfft()", "dims" is an optional subset of dimensions to
> > > transform, defaulting to "1:ndims(Afft)".
> > >
> > > "d" is the length of the transformed real array along the
> > > "dims[1]" dimension, which must satisfy "size(Afft,dims[1]) ==
> > > floor(d/2)+1". (This parameter cannot be inferred
> > > from "size(Afft)" due to the possibility of rounding by the
> > > "floor" function here.)
>
>
