一個程式來比較cuda/c在GPU/CPU的執行效率
阿新 • • 發佈:2019-01-30
在網上看了一個比較GPU和CPU執行矩陣運算效率的帖子,親自跑了一下。
這是一個CUDA語言程式,請儲存為“檔名.cu”。我的主機配置如下:
CPU:G2030;記憶體8GB;顯示卡:GTX750ti。
程式碼如下所示:
程式執行結果如下圖所示:#include "cuda_runtime.h" #include "device_launch_parameters.h" #include <stdio.h> #include <time.h> #define N (1024*1024) #define M (10000) #define THREADS_PER_BLOCK 1024 void serial_add(double *a, double *b, double *c, int n, int m) { for(int index=0;index<n;index++) { for(int j=0;j<m;j++) { c[index] = a[index]*a[index] + b[index]*b[index]; } } } __global__ void vector_add(double *a, double *b, double *c) { int index = blockIdx.x * blockDim.x + threadIdx.x; for(int j=0;j<M;j++) { c[index] = a[index]*a[index] + b[index]*b[index]; } } int main() { clock_t start,end; double *a, *b, *c; int size = N * sizeof( double ); a = (double *)malloc( size ); b = (double *)malloc( size ); c = (double *)malloc( size ); for( int i = 0; i < N; i++ ) { a[i] = b[i] = i; c[i] = 0; } start = clock(); serial_add(a, b, c, N, M); printf( "c[%d] = %f\n",0,c[0] ); printf( "c[%d] = %f\n",N-1, c[N-1] ); end = clock(); float time1 = ((float)(end-start))/CLOCKS_PER_SEC; printf("CPU: %f seconds\n",time1); start = clock(); double *d_a, *d_b, *d_c; cudaMalloc( (void **) &d_a, size ); cudaMalloc( (void **) &d_b, size ); cudaMalloc( (void **) &d_c, size ); cudaMemcpy( d_a, a, size, cudaMemcpyHostToDevice ); cudaMemcpy( d_b, b, size, cudaMemcpyHostToDevice ); vector_add<<< (N + (THREADS_PER_BLOCK-1)) / THREADS_PER_BLOCK, THREADS_PER_BLOCK >>>( d_a, d_b, d_c ); cudaMemcpy( c, d_c, size, cudaMemcpyDeviceToHost ); printf( "c[%d] = %f\n",0,c[0] ); printf( "c[%d] = %f\n",N-1, c[N-1] ); free(a); free(b); free(c); cudaFree( d_a ); cudaFree( d_b ); cudaFree( d_c ); end = clock(); float time2 = ((float)(end-start))/CLOCKS_PER_SEC; printf("CUDA: %f seconds, Speedup: %f\n",time2, time1/time2); return 0; }
CPU的執行時間是GPU執行時間的10倍,這已經是一個數量級的差距了。而且,我相信隨著運算量的加大,差距會更加明顯。由此看來GPU做矩陣運算確實比CPU快太多。
原帖地址:https://my.oschina.net/zzw922cn/blog/631650