float4 linear array or cudaArray?

Accelerated Computing CUDA CUDA Programming and Performance
1

Hi,

I would like to build a structure with 4 values. As I saw CUDA already provides the float4 type which allows you to save 4 float values and ensures fast memory access. However I have problems to define an array of float4 values.

I tried 2 versions, in the first one I just put the elements in a linear array, not knowing if CUDA handles a float4 just as one data element.

float4 weightedVoxel;

size_t size=100*sizeof(float4);

cudaMalloc((void**) &weightedVoxel, size);

weightedVoxel[1].x=1;

I get following error message: no operator “” matches these operands

operand types are: float4 [ int ]

However the acces to the element doesn’t work, so I tried the CUDA array, even I just have a 1D array of float 4 elements.

cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc<float4>();

cudaArray* weightedVoxel;

cudaMallocArray(&weightedVoxel, &channelDesc, 100,1);

weightedVoxel[1].x=1;

However thee I get the error message: ‘expression must be a pointer to a complete object type’

So, what I am doing wrong here? Is there a closer description of the different variable types, the one in the programming guide is really short?

2

make weightedVoxel a pointer and you’re all set

float4 *weightedVoxel;

size_t size=100*sizeof(float4);

cudaMalloc((void**) &weightedVoxel, size);

weightedVoxel[1].x=1;
3

[quote name=‘cbuchner1’ date=‘Sep 7 2008, 12:41 PM’]

make weightedVoxel a pointer and you’re all set

Thanks, I should have seen that.

It works now!

4

edit: nevermind