深度学习部署(十六): CUDA RunTime API _vector-add 使用cuda核函数实现向量加法
1. 知识点
- nthreads的取值,不能大于block能取值的最大值。一般可以直接给512、256,性能就是比较不错的
- (input_size + block_size - 1) / block_size;是向上取整
- 对于一维数组时,采用只定义layout的x维度,若处理的是二维,则可以考虑定义x、y维度,例如处理的是图像
- 关于把数据视作一维时,索引的计算
- 以下是通用的计算公式
Pseudo code: position = 0 for i in range(6):position *= dims[i]position += indexs[i]
- 例如当只使用x维度时,实际上dims = [1, 1, gd, 1, 1, bd],indexs = [0, 0, bi, 0, 0, ti]
- 因为0和1的存在,上面的循环则可以简化为:idx = threadIdx.x + blockIdx.x * blockDim.x
- 即:idx = ti + bi * bd
2. main.cpp文件
#include <cuda_runtime.h>
#include <stdio.h>#define checkRuntime(op) __check_cuda_runtime((op), #op, __FILE__, __LINE__)bool __check_cuda_runtime(cudaError_t code, const char* op, const char* file, int line){if(code != cudaSuccess){ const char* err_name = cudaGetErrorName(code); const char* err_message = cudaGetErrorString(code); printf("runtime error %s:%d %s failed. \n code = %s, message = %s\n", file, line, op, err_name, err_message); return false;}return true;
}void vector_add(const float* a, const float* b, float* c, int ndata);int main(){const int size = 3;float vector_a[size] = {2, 3, 2};float vector_b[size] = {5, 3, 3};float vector_c[size] = {0};float* vector_a_device = nullptr;float* vector_b_device = nullptr;float* vector_c_device = nullptr;checkRuntime(cudaMalloc(&vector_a_device, size * sizeof(float)));checkRuntime(cudaMalloc(&vector_b_device, size * sizeof(float)));checkRuntime(cudaMalloc(&vector_c_device, size * sizeof(float)));checkRuntime(cudaMemcpy(vector_a_device, vector_a, size * sizeof(float), cudaMemcpyHostToDevice));checkRuntime(cudaMemcpy(vector_b_device, vector_b, size * sizeof(float), cudaMemcpyHostToDevice));vector_add(vector_a_device, vector_b_device, vector_c_device, size);checkRuntime(cudaMemcpy(vector_c, vector_c_device, size * sizeof(float), cudaMemcpyDeviceToHost));for(int i = 0; i < size; ++i){printf("vector_c[%d] = %f\n", i, vector_c[i]);}checkRuntime(cudaFree(vector_a_device));checkRuntime(cudaFree(vector_b_device));checkRuntime(cudaFree(vector_c_device));return 0;
}
先定义三个数组: a, b, c 再用cudaMalloc()在GPU上开辟三个内存,在GPU上让a + b 并且让结果存储进c上,再把c的内存从GPU上放到Host上输出
3. 案例.cu文件
#include <stdio.h>
#include <cuda_runtime.h>__global__ void vector_add_kernel(const float* a, const float* b, float* c, int ndata){int idx = threadIdx.x + blockIdx.x * blockDim.x;if(idx >= ndata) return;/* dims indexsgridDim.z blockIdx.zgridDim.y blockIdx.ygridDim.x blockIdx.xblockDim.z threadIdx.zblockDim.y threadIdx.yblockDim.x threadIdx.xPseudo code:position = 0for i in 6:position *= dims[i]position += indexs[i]*/c[idx] = a[idx] + b[idx];
}void vector_add(const float* a, const float* b, float* c, int ndata){const int nthreads = 512;int block_size = ndata < nthreads ? ndata : nthreads; // 如果ndata < nthreads 那block_size = ndata就够了int grid_size = (ndata + block_size - 1) / block_size; // 其含义是我需要多少个blocks可以处理完所有的任务printf("block_size = %d, grid_size = %d\n", block_size, grid_size);vector_add_kernel<<<grid_size, block_size, 0, nullptr>>>(a, b, c, ndata);// 在核函数执行结束后,通过cudaPeekAtLastError获取得到的代码,来知道是否出现错误// cudaPeekAtLastError和cudaGetLastError都可以获取得到错误代码// cudaGetLastError是获取错误代码并清除掉,也就是再一次执行cudaGetLastError获取的会是success// 而cudaPeekAtLastError是获取当前错误,但是再一次执行cudaPeekAtLastError或者cudaGetLastErro拿到的还是那个错cudaError_t code = cudaPeekAtLastError();if(code != cudaSuccess){ const char* err_name = cudaGetErrorName(code); const char* err_message = cudaGetErrorString(code); printf("kernel error %s:%d test_print_kernel failed. \n code = %s, message = %s\n", __FILE__, __LINE__, err_name, err_message); }
}
两个注意的点
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像这个案例他就三个数相加,其实启动三个线程就足够了,但是一般block给的是512, 256,所以要设定一下,如果数组的长度小于256/512, 就直接用数组的长度的线程数就好。这里就是3个线程
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如果线程索引大于了数组的长度就直接返回了,不然就访问了不知道在哪里的内存了