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Copying data to "cufftComplex" data struct?

Tags:

cuda

memcpy

I have data stored as arrays of floats (single precision). I have one array for my real data, and one array for my complex data, which I use as the input to FFTs. I need to copy this data into the cufftComplex data type if I want to use the CUDA cufft library. From nVidia: " cufftComplex is a single‐precision, floating‐point complex data type that consists of interleaved real and imaginary components." Data to be operated on by cufft is stored in arrays of cufftComplex.

How do I quickly copy my data from a normal C array into an array of cufftComplex ? I don't want to use a for loop because it's probably the slowest possible option. I don't know how to use memcpy on arrays data of this type, because I do not know how it is stored in memory. Thanks!

like image 767
phasedweasel Avatar asked Nov 23 '12 20:11

phasedweasel


1 Answers

You could do this as part of a host-> device copy. Each copy would take one of the contiguous input arrays on the host and copy it in strided fashion to the device. The storage layout of complex data types in CUDA is compatible with the layout defined for complex types in Fortran and C++, i.e. as a structure with the real part followed by imaginary part.

float * real_vec;       // host vector, real part
float * imag_vec;       // host vector, imaginary part
float2 * complex_vec_d; // device vector, single-precision complex

float * tmp_d = (float *) complex_vec_d;

cudaStat = cudaMemcpy2D (tmp_d, 2 * sizeof(tmp_d[0]), 
                         real_vec, 1 * sizeof(real_vec[0]),
                         sizeof(real_vec[0]), n, cudaMemcpyHostToDevice);
cudaStat = cudaMemcpy2D (tmp_d + 1, 2 * sizeof(tmp_d[0]),
                         imag_vec, 1 * sizeof(imag_vec[0]),
                         sizeof(imag_vec[0]), n, cudaMemcpyHostToDevice);
like image 133
njuffa Avatar answered Nov 06 '22 16:11

njuffa