CUDA中内存访问越界问题
在核函数中对非法的内存访问不一定会报错,那我们加CHECK(error)会帮我们报错吗?经实验发现,有的非法访问很致命,函数不会运行,这种错误,可以CHECK到,但有些非法内存访问不致命,可能是访问到其他设备内存上了,这时CHECK不会报错,但是会出现逻辑上的错误,因为访问的值是错误的。
所以,我们在做内存访问时,一定要自己做好逻辑上的判断,保证内存访问不错位,不错误!
当x先申请设备内存的时候,y+1报错,x+1不报错,但属于逻辑错位。
当y先申请设备内存时,y+1逻辑错位,x+1报错。
例子:
#include <iostream>
#include <stdio.h>
#define SIZE 5
#define CHECK(res) { if(res != cudaSuccess){printf("Error :%s:%d , ", __FILE__,__LINE__); \
printf("code : %d , reason : %s \n", res,cudaGetErrorString(res));exit(-1);}}
/*
* 在内存交换的情况下,CPU 0.09 GPU 0.25
*/
#include <stdio.h>
#include <memory>
#include <math.h>
__global__
void saxpy(int n, float a, float *x, float *y)
{
int i = blockIdx.x*blockDim.x + threadIdx.x;
if(i<4)
{
printf("x: %f y :%f\n",x[i+2],y[i+1]);
}
y[i] = x[i] + y[i+1];
if(y[i]!=3){
printf("asd:x: %f y :%f\n",x[i],y[i]);
}
}
void add1(int n, float *x, float *y)
{
int i = 0;
for (i; i < n; i++)
{
//y[i] = sqrt(x[i] * y[i] * 5 * 5 * 95);
y[i] = x[i] + y[i];
}
}
int main(void)
{
int N =1<<20;
float *x, *y, *d_x, *d_y;
x = (float*)malloc(N * sizeof(float));
y = (float*)malloc(N * sizeof(float));
cudaMalloc(&d_y, N * sizeof(float));
cudaMalloc(&d_x, N * sizeof(float));
for (int i = 0; i < N; i++) {
x[i] = 1.0f;
y[i] = 2.0f;
}
cudaEvent_t start, stop;
cudaEventCreate(&start);
cudaEventCreate(&stop);
cudaMemcpy(d_x, x, N * sizeof(float), cudaMemcpyHostToDevice);
cudaMemcpy(d_y, y, N * sizeof(float), cudaMemcpyHostToDevice);
cudaEventRecord(start);
// Perform SAXPY on 1M elements
saxpy << <(N + 511) / 512, 512 >> >(N, 2.0f, d_x, d_y);
//add1(N,x,y);
CHECK(cudaDeviceSynchronize());
cudaEventRecord(stop);
cudaMemcpy(y, d_y, N * sizeof(float), cudaMemcpyDeviceToHost);
cudaEventSynchronize(stop);
float milliseconds = 0;
cudaEventElapsedTime(&milliseconds, start, stop);
float maxError = 0.0f;
for (int i = 0; i < N; i++) {
maxError = max(maxError, abs(y[i] - 3.0f));
}
printf("Max error: %f \n", maxError);
printf("Effective Bandwidth (GB/s): %f \n", milliseconds);
CHECK(cudaGetLastError());
cudaFree(d_x);
cudaFree(d_y);
free(x);
free(y);
}