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)?
- 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.)
>
>
>
