我有以下矩阵乘法代码，使用CUDA 3.2和VS 2008实现。我在Windows Server 2008 R2企业版上运行。我正在运行Nvidia GTX 480。以下代码适用于“宽度”（矩阵宽度）值为2500左右。
int size = Width*Width*sizeof(float);
float* Md, *Nd, *Pd;
cudaError_t err = cudaSuccess;
//Allocate Device Memory for M, N and P
err = cudaMalloc((void**)&Md, size);
err = cudaMalloc((void**)&Nd, size);
err = cudaMalloc((void**)&Pd, size);
//Copy Matrix from Host Memory to Device Memory
err = cudaMemcpy(Md, M, size, cudaMemcpyHostToDevice);
err = cudaMemcpy(Nd, N, size, cudaMemcpyHostToDevice);
//Setup the execution configuration
dim3 dimBlock(TileWidth, TileWidth, 1);
dim3 dimGrid(ceil((float)(Width)/TileWidth), ceil((float)(Width)/TileWidth), 1);
MatrixMultiplicationMultiBlock_Kernel<<<dimGrid, dimBlock>>>(Md, Nd, Pd, Width);
err = cudaMemcpy(P, Pd, size, cudaMemcpyDeviceToHost);
//Free Device Memory
cudaFree(Md);
cudaFree(Nd);
cudaFree(Pd);
当我将“Width”设置为3000或更大时，黑屏后出现以下错误：
我在网上查找后发现，有些人之所以会遇到这个问题，是因为看门狗在挂起超过5秒后会杀死内核。我尝试编辑注册表中的“TdrDelay”，这样可以延迟黑屏时间，但同样的错误也会出现。因此，我得出结论这不是我的问题。我对我的代码进行了调试，并发现以下一行是罪魁祸首：
err = cudaMemcpy(P, Pd, size, cudaMemcpyDeviceToHost);
这是我用来返回设备中矩阵乘法核函数调用后结果集的代码。在此之前，一切似乎都运行正常。我相信我正确地分配了内存，但是无法弄清楚为什么会出现这种情况。我认为可能是我的显卡上没有足够的内存，但是难道不应该在cudaMalloc时返回错误吗？（我在调试过程中确认它没有返回错误）。
有任何想法/帮助都将不胜感激！...非常感谢大家！！
核心代码：
//Matrix Multiplication Kernel - Multi-Block Implementation
__global__ void MatrixMultiplicationMultiBlock_Kernel (float* Md, float* Nd, float* Pd, int Width)
{
int TileWidth = blockDim.x;
//Get row and column from block and thread ids
int Row = (TileWidth*blockIdx.y) + threadIdx.y;
int Column = (TileWidth*blockIdx.x) + threadIdx.x;
//Pvalue store the Pd element that is computed by the thread
float Pvalue = 0;
for (int i = 0; i < Width; ++i)
{
float Mdelement = Md[Row * Width + i];
float Ndelement = Nd[i * Width + Column];
Pvalue += Mdelement * Ndelement;
}
//Write the matrix to device memory each thread writes one element
Pd[Row * Width + Column] = Pvalue;
}
我也有另一个使用共享内存的函数，它也会出现相同的错误：
调用：
MatrixMultiplicationSharedMemory_Kernel<<<dimGrid, dimBlock, sizeof(float)*TileWidth*TileWidth*2>>>(Md, Nd, Pd, Width);
内核代码：
//Matrix Multiplication Kernel - Shared Memory Implementation
__global__ void MatrixMultiplicationSharedMemory_Kernel (float* Md, float* Nd, float* Pd, int Width)
{
int TileWidth = blockDim.x;
//Initialize shared memory
extern __shared__ float sharedArrays[];
float* Mds = (float*) &sharedArrays;
float* Nds = (float*) &Mds[TileWidth*TileWidth];
int tx = threadIdx.x;
int ty = threadIdx.y;
//Get row and column from block and thread ids
int Row = (TileWidth*blockIdx.y) + ty;
int Column = (TileWidth*blockIdx.x) + tx;
float Pvalue = 0;
//For each tile, load the element into shared memory
for( int i = 0; i < ceil((float)Width/TileWidth); ++i)
{
Mds[ty*TileWidth+tx] = Md[Row*Width + (i*TileWidth + tx)];
Nds[ty*TileWidth+tx] = Nd[(ty + (i * TileWidth))*Width + Column];
__syncthreads();
for( int j = 0; j < TileWidth; ++j)
{
Pvalue += Mds[ty*TileWidth+j] * Nds[j*TileWidth+tx];
}
__syncthreads();
}
//Write the matrix to device memory each thread writes one element
Pd[Row * Width + Column] = Pvalue;
}
cuda-memcheck，对此感到抱歉。它应该能立即检测到越界访问。 - Tom