第一种方法：
CUDA thread index:
int blockId = blockIdx.z * (gridDim.x*gridDim.y)
+ blockIdx.y * gridDim.x
+ blockIdx.x;
int threadId = blockId * (blockDim.x * blockDim.y * blockDim.z)
+ threadIdx.z * (blockDim.x * blockDim.y)
+ threadIdx.y * blockDim.x
+ threadIdx.x;
从上到下分别是：
block的3,2,1维；
thread的3,2,1维。
如果没有对应维度，删除对应的计算部分即可。
维度的值最小是1，但是索引的值最小是0.
这里定义的格式为:
dim3 blocks(blockDim.x,blockDim.y,blockDim.z);
dim3 grids(gridDim.x,gridDim.y,gridDim.z); // 这里需要说明的是 如果的你的数组满足 M[NX][NY][NZ] 的形似，在设置blocks 和 grids时候 应该满足 blockDim.x * gridDim.x=NX, blockDim.y * gridDim.y =NY, blockDim.z * gridDim.z=NZ;
ps: 最少的情况是都只有1维，最多的情况是都有3维。
所以最简单的都只有1维时：blockIdx.x * blockDim.x + threadIdx.x
第二种方法：
分别计算 idx,idy,idz
idx = threadIdx.x + blockDim.x * blockIdx.x;
idy = threadIdx.y + blockDim.y * blockIdx.y;
idz = threadIdx.z + blockDim.z * blockIdx.z;
总的index (一维情况下)
id = idx + NX * idy + NX * NY * idz;
上面两种方法计算的结果是一样的，但是需要注意的是，第一种方法只能在index在一维情况下使用，而如果 index是二维的情况下的话，建议使用第二种计算方法，此时在设备端数组可以写成
M_dev[idx][idy][idz];
但是个人建议还是使用第一种方法，因为第二种方法涉及到二级指针问题，处理起来稍微会麻烦一点。
参考：https://www.cnblogs.com/tiandsp/p/9458734.html 有15种检索方式
#include "cuda_runtime.h" #include "device_launch_parameters.h" #include <stdio.h> #include <stdlib.h> #include <iostream> using namespace std; //thread 1D __global__ void testThread1(int *c, const int *a, const int *b) { int i = threadIdx.x; c[i] = b[i] - a[i]; } //thread 2D __global__ void testThread2(int *c, const int *a, const int *b) { int i = threadIdx.x + threadIdx.y*blockDim.x; c[i] = b[i] - a[i]; } //thread 3D __global__ void testThread3(int *c, const int *a, const int *b) { int i = threadIdx.x + threadIdx.y*blockDim.x + threadIdx.z*blockDim.x*blockDim.y; c[i] = b[i] - a[i]; } //block 1D __global__ void testBlock1(int *c, const int *a, const int *b) { int i = blockIdx.x; c[i] = b[i] - a[i]; } //block 2D __global__ void testBlock2(int *c, const int *a, const int *b) { int i = blockIdx.x + blockIdx.y*gridDim.x; c[i] = b[i] - a[i]; } //block 3D __global__ void testBlock3(int *c, const int *a, const int *b) { int i = blockIdx.x + blockIdx.y*gridDim.x + blockIdx.z*gridDim.x*gridDim.y; c[i] = b[i] - a[i]; } //block-thread 1D-1D __global__ void testBlockThread1(int *c, const int *a, const int *b) { int i = threadIdx.x + blockDim.x*blockIdx.x; c[i] = b[i] - a[i]; } //block-thread 1D-2D __global__ void testBlockThread2(int *c, const int *a, const int *b) { int threadId_2D = threadIdx.x + threadIdx.y*blockDim.x; int i = threadId_2D+ (blockDim.x*blockDim.y)*blockIdx.x; c[i] = b[i] - a[i]; } //block-thread 1D-3D __global__ void testBlockThread3(int *c, const int *a, const int *b) { int threadId_3D = threadIdx.x + threadIdx.y*blockDim.x + threadIdx.z*blockDim.x*blockDim.y; int i = threadId_3D + (blockDim.x*blockDim.y*blockDim.z)*blockIdx.x; c[i] = b[i] - a[i]; } //block-thread 2D-1D __global__ void testBlockThread4(int *c, const int *a, const int *b) { int blockId_2D = blockIdx.x + blockIdx.y*gridDim.x; int i = threadIdx.x + blockDim.x*blockId_2D; c[i] = b[i] - a[i]; } //block-thread 3D-1D __global__ void testBlockThread5(int *c, const int *a, const int *b) { int blockId_3D = blockIdx.x + blockIdx.y*gridDim.x + blockIdx.z*gridDim.x*gridDim.y; int i = threadIdx.x + blockDim.x*blockId_3D; c[i] = b[i] - a[i]; } //block-thread 2D-2D __global__ void testBlockThread6(int *c, const int *a, const int *b) { int threadId_2D = threadIdx.x + threadIdx.y*blockDim.x; int blockId_2D = blockIdx.x + blockIdx.y*gridDim.x; int i = threadId_2D + (blockDim.x*blockDim.y)*blockId_2D; c[i] = b[i] - a[i]; } //block-thread 2D-3D __global__ void testBlockThread7(int *c, const int *a, const int *b) { int threadId_3D = threadIdx.x + threadIdx.y*blockDim.x + threadIdx.z*blockDim.x*blockDim.y; int blockId_2D = blockIdx.x + blockIdx.y*gridDim.x; int i = threadId_3D + (blockDim.x*blockDim.y*blockDim.z)*blockId_2D; c[i] = b[i] - a[i]; } //block-thread 3D-2D __global__ void testBlockThread8(int *c, const int *a, const int *b) { int threadId_2D = threadIdx.x + threadIdx.y*blockDim.x; int blockId_3D = blockIdx.x + blockIdx.y*gridDim.x + blockIdx.z*gridDim.x*gridDim.y; int i = threadId_2D + (blockDim.x*blockDim.y)*blockId_3D; c[i] = b[i] - a[i]; } //block-thread 3D-3D __global__ void testBlockThread9(int *c, const int *a, const int *b) { int threadId_3D = threadIdx.x + threadIdx.y*blockDim.x + threadIdx.z*blockDim.x*blockDim.y; int blockId_3D = blockIdx.x + blockIdx.y*gridDim.x + blockIdx.z*gridDim.x*gridDim.y; int i = threadId_3D + (blockDim.x*blockDim.y*blockDim.z)*blockId_3D; c[i] = b[i] - a[i]; } void addWithCuda(int *c, const int *a, const int *b, unsigned int size) { int *dev_a = 0; int *dev_b = 0; int *dev_c = 0; cudaSetDevice(0); cudaMalloc((void**)&dev_c, size * sizeof(int)); cudaMalloc((void**)&dev_a, size * sizeof(int)); cudaMalloc((void**)&dev_b, size * sizeof(int)); cudaMemcpy(dev_a, a, size * sizeof(int), cudaMemcpyHostToDevice); cudaMemcpy(dev_b, b, size * sizeof(int), cudaMemcpyHostToDevice); //testThread1<<<1, size>>>(dev_c, dev_a, dev_b); //uint3 s;s.x = size/5;s.y = 5;s.z = 1; //testThread2 <<<1,s>>>(dev_c, dev_a, dev_b); //uint3 s; s.x = size / 10; s.y = 5; s.z = 2; //testThread3<<<1, s >>>(dev_c, dev_a, dev_b); //testBlock1<<<size,1 >>>(dev_c, dev_a, dev_b); //uint3 s; s.x = size / 5; s.y = 5; s.z = 1; //testBlock2<<<s, 1 >>>(dev_c, dev_a, dev_b); //uint3 s; s.x = size / 10; s.y = 5; s.z = 2; //testBlock3<<<s, 1 >>>(dev_c, dev_a, dev_b); //testBlockThread1<<<size/10, 10>>>(dev_c, dev_a, dev_b); //uint3 s1; s1.x = size / 100; s1.y = 1; s1.z = 1; //uint3 s2; s2.x = 10; s2.y = 10; s2.z = 1; //testBlockThread2 << <s1, s2 >> >(dev_c, dev_a, dev_b); //uint3 s1; s1.x = size / 100; s1.y = 1; s1.z = 1; //uint3 s2; s2.x = 10; s2.y = 5; s2.z = 2; //testBlockThread3 << <s1, s2 >> >(dev_c, dev_a, dev_b); //uint3 s1; s1.x = 10; s1.y = 10; s1.z = 1; //uint3 s2; s2.x = size / 100; s2.y = 1; s2.z = 1; //testBlockThread4 << <s1, s2 >> >(dev_c, dev_a, dev_b); //uint3 s1; s1.x = 10; s1.y = 5; s1.z = 2; //uint3 s2; s2.x = size / 100; s2.y = 1; s2.z = 1; //testBlockThread5 << <s1, s2 >> >(dev_c, dev_a, dev_b); //uint3 s1; s1.x = size / 100; s1.y = 10; s1.z = 1; //uint3 s2; s2.x = 5; s2.y = 2; s2.z = 1; //testBlockThread6 << <s1, s2 >> >(dev_c, dev_a, dev_b); //uint3 s1; s1.x = size / 100; s1.y = 5; s1.z = 1; //uint3 s2; s2.x = 5; s2.y = 2; s2.z = 2; //testBlockThread7 << <s1, s2 >> >(dev_c, dev_a, dev_b); //uint3 s1; s1.x = 5; s1.y = 2; s1.z = 2; //uint3 s2; s2.x = size / 100; s2.y = 5; s2.z = 1; //testBlockThread8 <<<s1, s2 >>>(dev_c, dev_a, dev_b); uint3 s1; s1.x = 5; s1.y = 2; s1.z = 2; uint3 s2; s2.x = size / 200; s2.y = 5; s2.z = 2; testBlockThread9<<<s1, s2 >>>(dev_c, dev_a, dev_b); cudaMemcpy(c, dev_c, size*sizeof(int), cudaMemcpyDeviceToHost); cudaFree(dev_a); cudaFree(dev_b); cudaFree(dev_c); cudaGetLastError(); } int main() { const int n = 1000; int *a = new int[n]; int *b = new int[n]; int *c = new int[n]; int *cc = new int[n]; for (int i = 0; i < n; i++) { a[i] = rand() % 100; b[i] = rand() % 100; c[i] = b[i] - a[i]; } addWithCuda(cc, a, b, n); FILE *fp = fopen("out.txt", "w"); for (int i = 0; i < n; i++) fprintf(fp, "%d %d\n", c[i], cc[i]); fclose(fp); bool flag = true; for (int i = 0; i < n; i++) { if (c[i] != cc[i]) { flag = false; break; } } if (flag == false) printf("no pass"); else printf("pass"); cudaDeviceReset(); delete[] a; delete[] b; delete[] c; delete[] cc; getchar(); return 0; }
其他补充内容，二维数组分块计算。
ix=threadIdx.x+blockDim.x*blockIdx.x iy=threadIdx.y+blockDim.y*blockIdx.y k=ix+iy*nx blockId =blockIdx.x+gridDim.x*blockIdx.y+gridDim.x*gridDim.y*blockIdx.z threadId=blockId*blockDim.x*blockIdDim.y*blockDim.z+ threadIdx.z*blockDim.x*blockDim.y threadIdx.y*blockDim.x threadIdx.x dim3 thread(tx ,ty ) dim3 block (NX/tx,NY/ty) gridDim.x=NX/tx gridDim.y=NY/ty blockDim.x=tx blockDim.y=ty a[ix][iy] 索引计算方法1： k=iy*NX+ix =(threadIdx.y+ty*blockIdx.y)*NX+tx*blockIdx.x +threadIdx.x =tx*blockIdx.x +NX*ty*blockIdx.y+NX*threadIdx.y+threadIdx.x 索引计算方法2： k=(blockIdx.x+NX/tx*blockIdx.y)*tx*ty+threadIdx.y*tx +threadIdx.x =tx*ty*blockIdx.x+NX*ty*blockIdx.y +tx*threadIdx.y +threadIdx.x
虽然上面两种计算方法的表达式不同，但是计算的结果都是一样的，索引值均是从0到NX*NY-1，总共NX*NY个坐标，证明如下，
dim3 thread(tx ,ty ) dim3 block (NX/tx,NY/ty) max{blockIdx.x} =NX/tx-1 max{blockIdx.y} =NY/ty-1 max{threadIdx.x}=tx-1 max{threadIdx.y}=ty-1 将上式带入到两种索引计算公式中，求得最大索引数 k = tx*blockIdx.x + NX*ty*blockIdx.y + NX*threadIdx.y + threadIdx.x = tx*(NX/tx -1) + NX*ty*(NY/ty-1) + NX*(ty-1) + tx-1 = NX*NY-1 k = tx*ty*blockIdx.x + NX*ty*blockIdx.y + tx*threadIdx.y + threadIdx.x = tx*ty*(NX/tx-1) + NX*ty*(NY/ty-1) + tx*(ty-1) + tx-1 = NX*NY-1
