Understanding CUDA grid dimensions, block dimensions and threads organization (simple explanation)
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;
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:
- func1 is executed for the first 512 threads
- func2 is executed for the first 512 threads
- func1 is executed for the remaining threads
- 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.
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.