right?īut why such a big 2D grid? I would have 256*4096 = 1,048,576 threads with that grid. Is that right?ĭim3 dimGrid( N/dimBlock.x, N/dimBlock.y ) // Means to me: N/dimBlock.x = 1024/16 = 64 and N/dimBlock.y = 64 → 64*64 = 4096 Blocks per grid. I do understand everything but not the give block and grid parameters.ĭim3 dimBlock( blocksize, blocksize ) // Means to me: 16*16 = 256 Threads per block. dim3 grid(Heightofinputimage/16, Width ofinputimage /16). Int j = blockIdx.y * blockDim.y + threadIdx.y ĬudaMemcpy( c, cd, size, cudaMemcpyDeviceToHost ) ĬudaFree( ad ) cudaFree( bd ) cudaFree( cd ) With the age of parallel computing technology, the sequential computation algorithms. Int i = blockIdx.x * blockDim.x + threadIdx.x _global_ void add_matrix( float* a, float *b, float *c, int N ) Int z_idx = blockIdx.z * blockDim.z + threadIdx.I am new to CUDA C GPGPU programming and found an example in the following pdf-file: Int y_idx = blockIdx.y * blockDim.y + threadIdx.y Int x_idx = blockIdx.x * blockDim.x + threadIdx.x _global_ void kernel(int xdim, int ydim, int zdim) Two input matrices of size Width x Width are M and N.

In this example, we will do the Square Matrix Multiplication. That is we must use two dimensional blocks. This feature in CUDA architecture enable us to create two-dimensional or even three-dimensional thread hierarchy so that solving two or three-dimensional problems becomes easier and more efficient. You are asked to calculate the index of this image pixels in the kernel function, but to run kernel the blocks are 16x16 sizes. The simplest approach, however, is to simply check for out-of-range. You have a kernel called 'ProcessImge(dprtIn, dptrOut)'. There are multiple ways to cope with this, such as ensuring that array dimensions are always a multiple of 16. Effectively your grid gets padded to a multiple of the block dimension. The downside is that now you may launch more threads than needed. This ensures that enough blocks are launched when the grid dimensions are not divisible by the block dimensions. This computation is a division rounding up.

The dim3type is equivalent to uint3with unspecified entries set to 1. When we call a kernel using the instruction <<< > we automatically define a dim3 type variable defining the number of blocks per grid and threads per. dim3 grid(1024, 768, 1) ĭim3 blocks((grid.x + blockdim.x - 1) / blockdim.x, CUDA Type dim3 CUDA uses the vector type dim3for the dimension variables, gridDimand blockDim. perform the calculation // setup execution parameters dim3 threads(4, 4) dim3 grid(1, 1) // execute the kernel matrixMul<<< grid, threads >(dC, dA.

Given this, you can calculate the blocks. When in doubt, stick with something small such as 128x1x1 or 16x16x1. This can lead to inefficient utilization because synchronization points such as _syncthreads() or the end of a kernel result in comparatibly many threads waiting for a few stragglers. If you have no specific need, I advise against using very large blocks such as 1024. Your block dimensions are dictated by your kernel. When defining a variable of type dim3, any component left. dim3 is an integer vector type based on uint3 that is used to specify dimensions. However, the access pattern depends on how you are interpreting your data and also how you are accessing them by 1D, 2D and 3D blocks of threads.

If a you have a couple of frames from a video, it might be dim3(width, height, frames). The memory is always a 1D continuous space of bytes. For-loop to calculate the value of result. Load these sub-matrices by block (sub-sub-matrices) of size (BLOCKSIZE, BLOCKSIZE). To review, open the file in an editor that reveals hidden Unicode characters. Dimensions of the grid in blocks (gridDim. Here we are going to have 1D grid with multiple thread blocks so we can calculate global index by summing up block offset for 1D thread block and threadIdx.x values. Then your grid would naturally be dim3(1024, 768, 1). Each kernel computes the result element (i,j). This file contains bidirectional Unicode text that may be interpreted or compiled differently than what appears below. This kernel takes 4 arguments - Pointer to the input vector A and B, Pointer to the output vector C and size of a vector. Let's say you have an image with dimensions 1024x768 and you do a pixel-wise computation. You start with defining your grid dimensions.

Dim3 grid calculation