Question about dimGrid

[juaninf]

Hi everyone

I am reading NVIDIA CUDA programming guide, and my question is when i write this :

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dim3 dimBlock(8);
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I am creating 8 threads? ie. an matrix de 1*8?

Other question i can access to block ID into of one grid with same example of the NVIDIA CUDA programming guide, by example if grid dimensions are Dx and Dy, if i want access to block (x,y) ;in the case two dimensions; a formule is (x+yDx)?

thanks

[ONeill]

Dim3 is just some datatype. But if you use this for your kernel launch configuration like kernel<<<dimGrid, dimBlock>>>() it will set the number of threads per block while dimGrid sets the number of blocks in the grid. So with kernel<<<1, 8>>>() you would have 8 threads, yes.
Your second thought is right, too.