【cuda入门系列】通过代码真实打印线程ID
- 1.`gridDim(6,1),blockDim(4,1)`
- 2.`gridDim(3,2),blockDim(2,2)`
【cuda入门系列之参加CUDA线上训练营】在Jetson nano本地跑 hello cuda！
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【cuda入门系列之参加CUDA线上训练营】共享内存实例1：矩阵转置实现及其优化
【cuda入门系列之参加CUDA线上训练营】共享内存实例2：矩阵相乘
【cuda入门系列】通过代码真实打印线程ID
定义一个长度为24的向量，分别用gridDim(6,1),blockDim(4,1)以及gridDim(3,2),blockDim(2,2)的thread去访问，确认thread与向量各元素之间的对应关系。
1.gridDim(6,1),blockDim(4,1)
#include <stdio.h>
#define BLOCK_SIZE 4
__global__ void gpu_print(int *a,int m,int n)
{
int row = blockIdx.y * blockDim.y + threadIdx.y;
int col = blockIdx.x * blockDim.x + threadIdx.x;
printf("%d %d\n", gridDim.x,gridDim.y);
printf("%d %d\n", blockDim.x,blockDim.y);
printf("blockIdx.y:%d blockIdx.x:%d threadIdx.y:%d threadIdx.x:%d val:%d \n", blockIdx.y,blockIdx.x,threadIdx.y,threadIdx.x,a[row*n+col]);
}
int main(int argc, char const *argv[])
{
int m=4;
int n=6;
int *h_a;
cudaMallocHost((void **) &h_a, sizeof(int)*m*n);
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
h_a[i * n + j] = i * n + j;
}
}
int *d_a;
cudaMalloc((void **) &d_a, sizeof(int)*m*n);
cudaMemcpy(d_a, h_a, sizeof(int)*m*n, cudaMemcpyHostToDevice);
dim3 dimGrid(6,1);
dim3 dimBlock(4,1);
gpu_print<<<dimGrid, dimBlock>>>(d_a,m, n);
// free memory
cudaFree(d_a);
cudaFreeHost(h_a);
system("pause");
return 0;
}
编译后打印结果如下：
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
6 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
4 1
blockIdx.y:0 blockIdx.x:1 threadIdx.y:0 threadIdx.x:0 val:4
blockIdx.y:0 blockIdx.x:1 threadIdx.y:0 threadIdx.x:1 val:5
blockIdx.y:0 blockIdx.x:1 threadIdx.y:0 threadIdx.x:2 val:6
blockIdx.y:0 blockIdx.x:1 threadIdx.y:0 threadIdx.x:3 val:7
blockIdx.y:0 blockIdx.x:3 threadIdx.y:0 threadIdx.x:0 val:12
blockIdx.y:0 blockIdx.x:3 threadIdx.y:0 threadIdx.x:1 val:13
blockIdx.y:0 blockIdx.x:3 threadIdx.y:0 threadIdx.x:2 val:14
blockIdx.y:0 blockIdx.x:3 threadIdx.y:0 threadIdx.x:3 val:15
blockIdx.y:0 blockIdx.x:2 threadIdx.y:0 threadIdx.x:0 val:8
blockIdx.y:0 blockIdx.x:2 threadIdx.y:0 threadIdx.x:1 val:9
blockIdx.y:0 blockIdx.x:2 threadIdx.y:0 threadIdx.x:2 val:10
blockIdx.y:0 blockIdx.x:2 threadIdx.y:0 threadIdx.x:3 val:11
blockIdx.y:0 blockIdx.x:4 threadIdx.y:0 threadIdx.x:0 val:16
blockIdx.y:0 blockIdx.x:4 threadIdx.y:0 threadIdx.x:1 val:17
blockIdx.y:0 blockIdx.x:4 threadIdx.y:0 threadIdx.x:2 val:18
blockIdx.y:0 blockIdx.x:4 threadIdx.y:0 threadIdx.x:3 val:19
blockIdx.y:0 blockIdx.x:0 threadIdx.y:0 threadIdx.x:0 val:0
blockIdx.y:0 blockIdx.x:0 threadIdx.y:0 threadIdx.x:1 val:1
blockIdx.y:0 blockIdx.x:0 threadIdx.y:0 threadIdx.x:2 val:2
blockIdx.y:0 blockIdx.x:0 threadIdx.y:0 threadIdx.x:3 val:3
blockIdx.y:0 blockIdx.x:5 threadIdx.y:0 threadIdx.x:0 val:20
blockIdx.y:0 blockIdx.x:5 threadIdx.y:0 threadIdx.x:1 val:21
blockIdx.y:0 blockIdx.x:5 threadIdx.y:0 threadIdx.x:2 val:22
blockIdx.y:0 blockIdx.x:5 threadIdx.y:0 threadIdx.x:3 val:23
从代码打印结果来看，一共有blcokDim4*gridDim 6=24个线程在工作。
- gridDim.x，gridDim.y———grid中x方向、y方向各含有多少个block;
- blockDim.x，blockDim.y——一个block中x方向、y方向各含有多少个thread。
定义的gridDim.x,gridDim.y以及blockDim.x,blockDim.y通过打印结果，可知：
各block中的thread与矩阵中元素的指向关系如下图：
2.gridDim(3,2),blockDim(2,2)
将代码中的
dim3 dimGrid(6,1);
dim3 dimBlock(4,1);
修改为：
dim3 dimGrid(3,2);
dim3 dimBlock(2,2);
其他不变，同样进行编译，打印输出：
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
3 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
2 2
blockIdx.y:0 blockIdx.x:1 threadIdx.y:0 threadIdx.x:0 val:2
blockIdx.y:0 blockIdx.x:1 threadIdx.y:0 threadIdx.x:1 val:3
blockIdx.y:0 blockIdx.x:1 threadIdx.y:1 threadIdx.x:0 val:8
blockIdx.y:0 blockIdx.x:1 threadIdx.y:1 threadIdx.x:1 val:9
blockIdx.y:1 blockIdx.x:0 threadIdx.y:0 threadIdx.x:0 val:12
blockIdx.y:1 blockIdx.x:0 threadIdx.y:0 threadIdx.x:1 val:13
blockIdx.y:1 blockIdx.x:0 threadIdx.y:1 threadIdx.x:0 val:18
blockIdx.y:1 blockIdx.x:0 threadIdx.y:1 threadIdx.x:1 val:19
blockIdx.y:0 blockIdx.x:2 threadIdx.y:0 threadIdx.x:0 val:4
blockIdx.y:0 blockIdx.x:2 threadIdx.y:0 threadIdx.x:1 val:5
blockIdx.y:0 blockIdx.x:2 threadIdx.y:1 threadIdx.x:0 val:10
blockIdx.y:0 blockIdx.x:2 threadIdx.y:1 threadIdx.x:1 val:11
blockIdx.y:1 blockIdx.x:1 threadIdx.y:0 threadIdx.x:0 val:14
blockIdx.y:1 blockIdx.x:1 threadIdx.y:0 threadIdx.x:1 val:15
blockIdx.y:1 blockIdx.x:1 threadIdx.y:1 threadIdx.x:0 val:20
blockIdx.y:1 blockIdx.x:1 threadIdx.y:1 threadIdx.x:1 val:21
blockIdx.y:0 blockIdx.x:0 threadIdx.y:0 threadIdx.x:0 val:0
blockIdx.y:0 blockIdx.x:0 threadIdx.y:0 threadIdx.x:1 val:1
blockIdx.y:0 blockIdx.x:0 threadIdx.y:1 threadIdx.x:0 val:6
blockIdx.y:0 blockIdx.x:0 threadIdx.y:1 threadIdx.x:1 val:7
blockIdx.y:1 blockIdx.x:2 threadIdx.y:0 threadIdx.x:0 val:16
blockIdx.y:1 blockIdx.x:2 threadIdx.y:0 threadIdx.x:1 val:17
blockIdx.y:1 blockIdx.x:2 threadIdx.y:1 threadIdx.x:0 val:22
blockIdx.y:1 blockIdx.x:2 threadIdx.y:1 threadIdx.x:1 val:23
貌似是先切割y方向，比如此例子中，gridDim.yblockDim.y=22=4，所以将24个元素平分成了4份；然后再在x方向分割。最后组装，由各block中的thread访问。
