我必须比较CUDA中的两个浮点阵列(a,b),以便
if a > b then a = a/a ; else a = 0
。
请告诉您调用此方法的正确方法和语法。
答案 0 :(得分:1)
这样的事情应该有效。为简洁起见,我正在简化我常用的cuda错误检查。
#include <stdio.h>
#define DSIZE 10000
#define nTPB 512
__global__ void cmp(float *a, float *b, int size){
int idx = threadIdx.x + blockDim.x*blockIdx.x;
if (idx < size)
a[idx]=(a[idx] > b[idx])?1.0f:0.0f; // could also be: ?(a[idx]/a[idx]):0;
}
int main() {
cudaError_t err;
float *h_a, *h_b, *d_a, *d_b;
h_a = (float *)malloc(DSIZE*sizeof(float));
if (h_a == 0) {printf("malloc fail\n"); return 1;}
h_b = (float *)malloc(DSIZE*sizeof(float));
if (h_b == 0) {printf("malloc fail\n"); return 1;}
for (int i=0; i< DSIZE; i++){
h_a[i] = 10.0f;
h_b[i] = (float)i;}
err = cudaMalloc((void **)&d_a, DSIZE*sizeof(float));
if (err != cudaSuccess) {printf("cuda fail\n"); return 1;}
err = cudaMalloc((void **)&d_b, DSIZE*sizeof(float));
if (err != cudaSuccess) {printf("cuda fail\n"); return 1;}
err = cudaMemcpy(d_a, h_a, DSIZE*sizeof(float), cudaMemcpyHostToDevice);
if (err != cudaSuccess) {printf("cuda fail\n"); return 1;}
err = cudaMemcpy(d_b, h_b, DSIZE*sizeof(float), cudaMemcpyHostToDevice);
if (err != cudaSuccess) {printf("cuda fail\n"); return 1;}
cmp<<<(DSIZE+nTPB-1)/nTPB, nTPB>>>(d_a, d_b, DSIZE);
err=cudaMemcpy(h_a, d_a, DSIZE*sizeof(float), cudaMemcpyDeviceToHost);
if (err != cudaSuccess) {printf("cuda fail\n"); return 1;}
for (int i=0; i< 20; i++)
printf("h_a[%d] = %f\n", i, h_a[i]);
return 0;
}