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The following line was working previously at version 5.4.3.. However, with v5.5.2, it fails to precompile:
function CUDA.cu(P::FFTW.rFFTWPlan)
return plan_rfft(cu(zeros(real(eltype(P)), P.sz)), P.region)
end
CUDA.cu(P::CUFFT.rCuFFTPlan) = P
The above generates the error: rCuFFTPlan not defined.
For v5.4.3 and earlier, we had to do the above since CUDA.cu did not accept FFTW.rFFTWPlan. For v5.5.2, does it accept FFTW.rFFTWPlan directly so that the above conversion function is not necessary anymore?
Thanks a lot for your help!
The text was updated successfully, but these errors were encountered:
That's doing both type piracy (you shouldn't extend the cu function, as mentioned in the docstring, and definitely not on types you do not own), while using internal types that are not guaranteed to be stable. CUFFT.rCuFFTPlan was removed in https://github.com/JuliaGPU/CUDA.jl/pull/2430/files because of switching to the new unified Xt interface; can you look at the changes in that PR to see if they fit your use case?
And for future reference: Instead of carrying code like that, it's better to open a PR to fix the underlying problem, or at the very least open an issue so that somebody else can work on it.
The following line was working previously at version 5.4.3.. However, with v5.5.2, it fails to precompile:
The above generates the error:
rCuFFTPlan
not defined.For v5.4.3 and earlier, we had to do the above since
CUDA.cu
did not acceptFFTW.rFFTWPlan
. For v5.5.2, does it acceptFFTW.rFFTWPlan
directly so that the above conversion function is not necessary anymore?Thanks a lot for your help!
The text was updated successfully, but these errors were encountered: