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我正在使用特斯拉k40c gpu。我在其上運行下面的代碼:在此代碼中使用共享內存會導致錯誤嗎?
#define gpuErrchk(ans) { gpuAssert((ans), __FILE__, __LINE__); }
inline void gpuAssert(cudaError_t code, const char *file, int line, bool abort=true)
{
if (code != cudaSuccess)
{
fprintf(stderr,"GPUassert: %s %s %d\n", cudaGetErrorString(code), file, line);
if (abort) exit(code);
}
}
__global__ void ker(float * a, float * c,long long cor_size,int n, int m)
{
extern __shared__ float cache[];
cache[threadIdx.x]=4;
}
int main()
{
int m =500,n =10000;
long long cor_size = n-1;
cor_size *=n;
cor_size /=2;
float * dev_bold1,*dev_bold3;
gpuErrchk(cudaMalloc ((void**)&dev_bold1,sizeof(float)*m*n));
gpuErrchk(cudaMalloc ((void**)&dev_bold3,sizeof(float)*cor_size));
float * bold1 = new float [m*n];
memset(bold1, 0, sizeof(float)*m*n);
float * bold3 = new float [cor_size];
memset(bold3, 0, sizeof(float) *cor_size);
gpuErrchk(cudaMemcpy(dev_bold1, bold1, sizeof(float) * m * n, cudaMemcpyHostToDevice));
ker<<<cor_size,m,m>>>(dev_bold1,dev_bold3,cor_size,n,m);
gpuErrchk(cudaPeekAtLastError());
gpuErrchk(cudaMemcpy(bold3, dev_bold3, sizeof(float)*cor_size, cudaMemcpyDeviceToHost));
return 1;
}
運行的代碼後,我得到了以下錯誤:
GPUassert: an illegal memory access was encountered test2.cu 48
48號線是指
gpuErrchk(cudaMemcpy(bold3, dev_bold3, sizeof(float)*cor_size, cudaMemcpyDeviceToHost));
我不明白是什麼原因造成的錯誤,但是當我將m的值更改爲50或更少的代碼工作,但對於更高的值,它會給我這個錯誤。這個問題與使用共享內存有關嗎?
是的,這是問題,謝謝! – starrr