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Why Does `cudaMemcpy` Cause a Segmentation Fault When Copying from Device to Host, and How Can I Fix It?

Patricia Arquette
Release: 2024-12-03 21:47:12
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Why Does `cudaMemcpy` Cause a Segmentation Fault When Copying from Device to Host, and How Can I Fix It?

cudaMemcpy Segmentation Fault

When calling cudaMemcpy to transfer data from device to host, users may encounter a segmentation fault. This often stems from improper handling of device pointers.

The Issue

The following code demonstrates the issue:

cudaMemcpy(CurrentGrid->cdata[i], Grid_dev->cdata[i], size*sizeof(float),\
                cudaMemcpyDeviceToHost);
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Here, Grid_dev is a device pointer to a grid class object. Attempting to directly dereference this pointer in the cudaMemcpy call may lead to a segmentation fault.

Solution

To resolve this issue, an intermediate pointer must be created on the host to reference the device pointer. The following code outlines the revised solution:

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);            
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In this modified code, A is a host pointer that initially points to a device pointer. By copying the value of Grid_dev->cdata[i] to A using cudaMemcpy, we effectively transfer the device pointer to the host. Subsequently, CurrentGrid->cdata[i] can be assigned to a new float array on the host, and the data can be copied from the device pointer A to this array.

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