【发布时间】:2014-06-14 14:29:56
【问题描述】:
我在将主机内存复制到设备内存后尝试访问设备内存。当尝试打印从主机内存复制到设备内存的数据时,执行结果不正确..它说分段错误,我知道我正在尝试从不可用的设备内存打印数据或无法访问。
帮助我如何访问此设备内存,并且我想确保如果我修改主机内存数据,那么我希望在尝试打印时在设备内存数据中看到该更改。
下面是我的代码
// includes, system
#include <stdio.h>
#include <assert.h>
// Simple utility function to check for CUDA runtime errors
void checkCUDAError(const char *msg);
int main( int argc, char** argv)
{
// pointer and dimension for host memory
int n, dimA;
float *h_a;
// pointers for device memory
float *d_a, *d_b;
// allocate and initialize host memory
/** Bonus: try using cudaMallocHost in place of malloc **/
dimA = 8;
size_t memSize = dimA*sizeof(float);
cudaMallocHost((void**)&h_a, memSize);
//h_a = (float *) malloc(dimA*sizeof(float));
for (n=0; n<dimA; n++)
{
h_a[n] = (float) n;
}
// Part 1 of 5: allocate device memory
cudaMalloc( (void**)&d_a, memSize );
cudaMalloc( (void**)&d_b, memSize );
// Part 2 of 5: host to device memory copy
cudaMemcpy( d_a, h_a, memSize, cudaMemcpyHostToDevice );
// Part 3 of 5: device to device memory copy
cudaMemcpy( d_b, d_a, memSize, cudaMemcpyDeviceToDevice );
// clear host memory
for (n=0; n<dimA; n++)
{
printf("Data in host memory h_a %f\n", h_a[n]);
printf("Data in device memory d_a %f\n", d_a[n]);
//printf("Data in device memory d_b %f\n", d_b[n]);
h_a[n] = 0.f;
}
// Part 4 of 5: device to host copy
cudaMemcpy( h_a, d_b, memSize, cudaMemcpyDeviceToHost );
// Check for any CUDA errors
checkCUDAError("cudaMemcpy calls");
// verify the data on the host is correct
for (n=0; n<dimA; n++)
{
assert(h_a[n] == (float) n);
}
// Part 5 of 5: free device memory pointers d_a and d_b
cudaFree( d_b );
cudaFree( d_a );
// Check for any CUDA errors
checkCUDAError("cudaFree");
// free host memory pointer h_a
// Bonus: be sure to use cudaFreeHost for memory allocated with cudaMallocHost
cudaFreeHost(h_a);
//free(h_a);
// If the program makes it this far, then the results are correct and
// there are no run-time errors. Good work!
printf("cudaMallocHost is working Correct!\n");
return 0;
}
void checkCUDAError(const char *msg)
{
cudaError_t err = cudaGetLastError();
if( cudaSuccess != err)
{
fprintf(stderr, "Cuda error: %s: %s.\n", msg, cudaGetErrorString( err) );
exit(-1);
}
}
所以在将内存从 d_a 复制到 d_b 之后的代码中,当我尝试在 d_b 内存中打印数据时,它会给出错误。并且打印 h_a 内存会产生很好的效果。我在尝试在 d_b 内存中打印数据时做错了吗?
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