CUFFT with double precision

Question

I am experiencing some problems with CUDAs FFT library.

I declared the inputs as cuDoubleComplex, but the compiler returns the error that this type is incompatible with parameters of type cufftComplex. After some search through the Internet, I found the file cufft.h, in which there is the line typedef cuComplex cufftComplex; . My problem is that in the library cuComplex.h it is clear that cuComplex has a single floating point precision ( typedef cuFloatComplex cuComplex; ), but I would like a double precision.

Is this possible?

In particular, I obtain the following:

error: argument of type "cufftDoubleComplex *" is incompatible with parameter of type "cufftComplex *"

at this line:

cufftExecC2C(plan, data1, data2, CUFFT_FORWARD);

Answer 1

The double precision complex data type is defined as cufftDoubleComplex in CUFFT.

Double precision versions of fft in CUFFT are:

cufftExecD2Z() //Real To Complex

cufftExecZ2D() //Complex To Real

cufftExecZ2Z() //Complex To Complex

cufftExecC2C is the single precision version of fft, and expects the input and output pointers to be of type cufftComplex ,whereas you are passing it a pointer of type cufftDoubleComplex .

For cufftDoubleComplex data type, you have to use the function cufftExecZ2Z instead, which is for double precision data.