Understanding CUDA grid dimensions, block dimensions and threads organization (simple explanation) [closed]

Answer 1

Hardware

If a GPU device has, for example, 4 multiprocessing units, and they can run 768 threads each: then at a given moment no more than 4*768 threads will be really running in parallel (if you planned more threads, they will be waiting their turn).

Software

threads are organized in blocks. A block is executed by a multiprocessing unit. The threads of a block can be indentified (indexed) using 1Dimension(x), 2Dimensions (x,y) or 3Dim indexes (x,y,z) but in any case xyz <= 768 for our example (other restrictions apply to x,y,z, see the guide and your device capability).

Obviously, if you need more than those 4*768 threads you need more than 4 blocks. Blocks may be also indexed 1D, 2D or 3D. There is a queue of blocks waiting to enter the GPU (because, in our example, the GPU has 4 multiprocessors and only 4 blocks are being executed simultaneously).

Now a simple case: processing a 512x512 image

Suppose we want one thread to process one pixel (i,j).

We can use blocks of 64 threads each. Then we need 512*512/64 = 4096 blocks (so to have 512x512 threads = 4096*64)

It's common to organize (to make indexing the image easier) the threads in 2D blocks having blockDim = 8 x 8 (the 64 threads per block). I prefer to call it threadsPerBlock.

dim3 threadsPerBlock(8, 8);  // 64 threads

and 2D gridDim = 64 x 64 blocks (the 4096 blocks needed). I prefer to call it numBlocks.

dim3 numBlocks(imageWidth/threadsPerBlock.x,  /* for instance 512/8 = 64*/
              imageHeight/threadsPerBlock.y); 

The kernel is launched like this:

myKernel <<<numBlocks,threadsPerBlock>>>( /* params for the kernel function */ );       

Finally: there will be something like "a queue of 4096 blocks", where a block is waiting to be assigned one of the multiprocessors of the GPU to get its 64 threads executed.

In the kernel the pixel (i,j) to be processed by a thread is calculated this way:

uint i = (blockIdx.x * blockDim.x) + threadIdx.x;
uint j = (blockIdx.y * blockDim.y) + threadIdx.y;

Answer 2

Suppose a 9800GT GPU:

• it has 14 multiprocessors (SM)
• each SM has 8 thread-processors (AKA stream-processors, SP or cores)
• allows up to 512 threads per block
• warpsize is 32 (which means each of the 14x8=112 thread-processors can schedule up to 32 threads)

https://www.tutorialspoint.com/cuda/cuda_threads.htm

A block cannot have more active threads than 512 therefore __syncthreads can only synchronize limited number of threads. i.e. If you execute the following with 600 threads:

func1();
__syncthreads();
func2();
__syncthreads();

then the kernel must run twice and the order of execution will be:

1. func1 is executed for the first 512 threads
2. func2 is executed for the first 512 threads
3. func1 is executed for the remaining threads
4. func2 is executed for the remaining threads

Note:

The main point is __syncthreads is a block-wide operation and it does not synchronize all threads.

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I'm not sure about the exact number of threads that __syncthreads can synchronize, since you can create a block with more than 512 threads and let the warp handle the scheduling. To my understanding it's more accurate to say: func1 is executed at least for the first 512 threads.

Before I edited this answer (back in 2010) I measured 14x8x32 threads were synchronized using __syncthreads.

I would greatly appreciate if someone test this again for a more accurate piece of information.