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Why Does cudaMemcpy Cause a Segmentation Fault When Dereferencing a NULL Device Pointer?

Linda Hamilton
Release: 2024-12-04 15:30:13
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Why Does cudaMemcpy Cause a Segmentation Fault When Dereferencing a NULL Device Pointer?

cudaMemcpy Segmentation Fault: Insight and Troubleshooting

The "cudaMemcpy segmentation fault" error often arises when cudaMemcpy operates on invalid memory addresses. To delve into this issue, let's focus on a specific example from the posted inquiry:

cudaMemcpy(CurrentGrid->cdata[i], Grid_dev->cdata[i], size*sizeof(float),\
                cudaMemcpyDeviceToHost);
Copy after login

Upon investigating the code and debug information, it was discovered that the pointer Grid_dev->cdata[i] was NULL on the device, which resulted in a segmentation fault when being dereferenced in the cudaMemcpy call.

Why Dereferencing Device Pointers Fails

While device pointers can be employed in cudaMemcpy calls, it's imperative to remember that the pointer only stores the device address. To access the actual data on the device, we need to perform an additional cudaMemcpy to copy the pointer value from the device to a host pointer. This host pointer can then be used to access the data.

Revised Code to Address This Issue

The original code has been amended with a more appropriate approach:

float * A;
cudaMalloc((void**)&A, sizeof(float));
...
...
cudaMemcpy(&A, &(Grid_dev->cdata[i]), sizeof(float *), cudaMemcpyDeviceToHost);    
CurrentGrid->cdata[i] = new float[size];
cudaMemcpy(CurrentGrid->cdata[i], A, size*sizeof(float), cudaMemcpyDeviceToHost);            
Copy after login

Here, we allocate a float pointer A on the device and cudaMemcpy the value of Grid_dev->cdata[i] to A. Then, we cudaMemcpy A to the host. This ensures that we capture the pointer value rather than attempting to dereference it directly.

Potential Memory Leak

The modified code may incur a memory leak if the pointer A is not freed after the data copy operation. To mitigate this, the memory allocated for A should be released using cudaFree(A).

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