CUDA编程--实现并行矩阵乘法【80行代码】

简述

这里只写了方阵之间的乘法，但是本质上都是一样的。

• 我测试过100规模的方阵之间的乘法，没有问题。

代码

• 读取文件data.txt
• 数据格式就是一个数值N，然后来连续的两个N*N的矩阵。用空格隔开。

#include "cuda_runtime.h" #include "device_launch_parameters.h" #include <iostream> #include <fstream> #include <stdio.h> // Kernal: __global__ void MatrixMultiply(int *a, int * b, int *c, int N) {int tx = threadIdx.x + blockIdx.x * blockDim.x;int ty = threadIdx.y + blockIdx.y * blockDim.y;if (tx < N && ty < N) {int sum = 0;for (int k = 0; k < N; ++k) {int adata = a[tx * N + k];int bdata = b[k * N + ty];sum += adata * bdata;}c[tx * N + ty] = sum;} }cudaError_t matrixMultiplyWithCuda(int *a, int *b, int *c, size_t size);int main() {std::ifstream in("data.txt");int N;in >> N;if (in.fail()) {printf("Something wrong\n");}else {printf("Success read\n");}// host initialint *a = new int[N * N];int *b = new int[N * N];int *c = new int[N * N];// read for (int i = 0; i < N; ++i)for (int j = 0; j < N; ++j) in >> a[i * N + j];for (int i = 0; i < N; ++i)for (int j = 0; j < N; ++j) in >> b[i * N + j];cudaError_t cudaStatus = matrixMultiplyWithCuda(a, b, c, N);for (int i = 0; i < N; ++i) {for (int j = 0; j < N; ++j) std::cout << c[i * N + j]<<" ";std::cout << std::endl;}cudaStatus = cudaThreadExit();// host free delete[] a;delete[] b;delete[] c;return 0; } cudaError_t matrixMultiplyWithCuda(int *a, int *b, int *c, size_t N) {int *dev_a = 0;int *dev_b = 0;int *dev_c = 0;cudaError_t cudaStatus;cudaStatus = cudaMalloc((void**)&dev_a, N * N * sizeof(int));cudaStatus = cudaMalloc((void**)&dev_b, N * N * sizeof(int));cudaStatus = cudaMalloc((void**)&dev_c, N * N * sizeof(int));cudaStatus = cudaMemcpy(dev_a, a, N * N * sizeof(int), cudaMemcpyHostToDevice);cudaStatus = cudaMemcpy(dev_b, b, N * N * sizeof(int), cudaMemcpyHostToDevice);if (cudaStatus != cudaSuccess) {printf("Something wrong\n");goto Error;}// kernal invocation dim3 threadPerBlock(32, 32);dim3 numBlocks(N / threadPerBlock.x + 1, N / threadPerBlock.y + 1);MatrixMultiply<<<numBlocks, threadPerBlock>>>(dev_a, dev_b, dev_c, N);if (cudaStatus != cudaSuccess) {printf( "Calculate wrong\n");goto Error;}cudaStatus = cudaMemcpy(c, dev_c, N * N * sizeof(int), cudaMemcpyDeviceToHost); Error:cudaFree(dev_a);cudaFree(dev_b);cudaFree(dev_c);return cudaStatus; }

写入文件的 版本

（也改成了浮点数运算了）

#include "cuda_runtime.h" #include "device_launch_parameters.h" #include <iostream> #include <fstream> #include <stdio.h> // Kernal: __global__ void MatrixMultiply(float *a, float * b, float *c, int N) {int tx = threadIdx.x + blockIdx.x * blockDim.x;int ty = threadIdx.y + blockIdx.y * blockDim.y;if (tx < N && ty < N) {float sum = 0;for (int k = 0; k < N; ++k) {float adata = a[tx * N + k];float bdata = b[k * N + ty];sum += adata * bdata;}c[tx * N + ty] = sum;} }cudaError_t matrixMultiplyWithCuda(float *a, float *b, float *c, size_t size);int main() {std::ifstream in("data.txt");int N;in >> N;if (in.fail()) {printf("Something wrong\n");}else {printf("Success read\n");}// host initialfloat *a = new float[N * N];float *b = new float[N * N];float *c = new float[N * N];// read for (int i = 0; i < N; ++i)for (int j = 0; j < N; ++j) in >> a[i * N + j];for (int i = 0; i < N; ++i)for (int j = 0; j < N; ++j) in >> b[i * N + j];cudaError_t cudaStatus = matrixMultiplyWithCuda(a, b, c, N);std::ofstream out("output.txt");for (int i = 0; i < N; ++i) {for (int j = 0; j < N; ++j) out << c[i * N + j]<<" ";out << std::endl;}cudaStatus = cudaThreadExit();// host free delete[] a;delete[] b;delete[] c;return 0; } cudaError_t matrixMultiplyWithCuda(float *a, float *b, float *c, size_t N) {float *dev_a = 0;float *dev_b = 0;float *dev_c = 0;cudaError_t cudaStatus;cudaStatus = cudaMalloc((void**)&dev_a, N * N * sizeof(int));cudaStatus = cudaMalloc((void**)&dev_b, N * N * sizeof(int));cudaStatus = cudaMalloc((void**)&dev_c, N * N * sizeof(int));cudaStatus = cudaMemcpy(dev_a, a, N * N * sizeof(int), cudaMemcpyHostToDevice);cudaStatus = cudaMemcpy(dev_b, b, N * N * sizeof(int), cudaMemcpyHostToDevice);if (cudaStatus != cudaSuccess) {printf("Something wrong\n");goto Error;}// kernal invocation dim3 threadPerBlock(32, 32);dim3 numBlocks(N / threadPerBlock.x + 1, N / threadPerBlock.y + 1);MatrixMultiply<<<numBlocks, threadPerBlock>>>(dev_a, dev_b, dev_c, N);if (cudaStatus != cudaSuccess) {printf( "Calculate wrong\n");goto Error;}cudaStatus = cudaMemcpy(c, dev_c, N * N * sizeof(int), cudaMemcpyDeviceToHost); Error:cudaFree(dev_a);cudaFree(dev_b);cudaFree(dev_c);return cudaStatus; }

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