问题
If I start my kernel with a grid whose blocks have dimensions:
dim3 block_dims(16,16);
How are the grid blocks now split into warps? Do the first two rows of such a block form one warp, or the first two columns, or is this arbitrarily-ordered?
Assume a GPU Compute Capability of 2.0.
回答1:
Threads are numbered in order within blocks so that threadIdx.x varies the fastest, then threadIdx.y the second fastest varying, and threadIdx.z the slowest varying. This is functionally the same as column major ordering in multidimensional arrays. Warps are sequentially constructed from threads in this ordering. So the calculation for a 2d block is
unsigned int tid = threadIdx.x + threadIdx.y * blockDim.x;
unsigned int warpid = tid / warpSize;
This is covered both in the programming guide and the PTX guide.
回答2:
To illustrate @talonmies's answer through 'Visual Studio WarpWatch' window for two consecutive warps (dim3 block_dims(16,16); and WarpSize = 32):
来源：https://stackoverflow.com/questions/6177202/how-are-cuda-blocks-divided-into-warps