How do I use cudaMemcpy2D() DeviceToHost - visual-c++

Question

I'm new to cuda and C++ and just can't seem to figure this out.
What I want to do is copy a 2d array A to the device then copy it back to an identical array B.
I would expect that the B array would have the same values as A, but there is something that I'm doing wrong.
CUDA - 4.2, compiling for win32, 64 bit machine, NVIDIA Quadro K5000
Here is the code.
void main(){
cout<<"Host main" << endl;
// Host code
const int width = 3;
const int height = 3;
float* devPtr;
float a[width][height];
//load and display input array
cout << "a array: "<< endl;
for (int i = 0 ; i < width; i ++)
{
for (int j = 0 ; j < height; j ++)
{
a[i][j] = i + j;
cout << a[i][j] << " ";
}
cout << endl;
}
cout<< endl;
//Allocating Device memory for 2D array using pitch
size_t host_orig_pitch = width * sizeof(float); //host original array pitch in bytes
size_t pitch;// pitch for the device array
cudaMallocPitch(&devPtr, &pitch, width * sizeof(float), height);
cout << "host_orig_pitch: " << host_orig_pitch << endl;
cout << "sizeof(float): " << sizeof(float)<< endl;
cout << "width: " << width << endl;
cout << "height: " << height << endl;
cout << "pitch: " << pitch << endl;
cout << endl;
cudaMemcpy2D(devPtr, pitch, a, host_orig_pitch, width, height, cudaMemcpyHostToDevice);
float b[width][height];
//load b and display array
cout << "b array: "<< endl;
for (int i = 0 ; i < width; i ++)
{
for (int j = 0 ; j < height; j ++)
{
b[i][j] = 0;
cout << b[i][j] << " ";
}
cout << endl;
}
cout<< endl;
//MyKernel<<<100, 512>>>(devPtr, pitch, width, height);
//cudaThreadSynchronize();
//cudaMemcpy2d(dst, dPitch,src ,sPitch, width, height, typeOfCopy )
cudaMemcpy2D(b, host_orig_pitch, devPtr, pitch, width, height, cudaMemcpyDeviceToHost);
// should be filled in with the values of array a.
cout << "returned array" << endl;
for(int i = 0 ; i < width ; i++){
for (int j = 0 ; j < height ; j++){
cout<< b[i][j] << " " ;
}
cout<<endl;
}
cout<<endl;
system("pause");
}
Here is the output.
Host main A Array 0 1 2 1 2 3 2 3 4
host_orig_pitch: 12 sizeof(float): 4 width: 3 height: 3 pitch: 512
b array: 0 0 0 0 0 0 0 0 0
returned array 0 0 0
1.17549e-038 0 0 0 0 0
Press any key to continue . . .
If more information is need let me know and I'll post it.
Any help would be greatly appreciated.

Answer 1

As identified in comments, the original poster was supplying incorrect arguments to the cudaMemcpy2Dcall. The width argument for a transfer is always in bytes, so in the above code:
cudaMemcpy2D(b, host_orig_pitch, devPtr, pitch, width, height, cudaMemcpyDeviceToHost);
should be
cudaMemcpy2D(b, host_orig_pitch, devPtr, pitch, width * sizeof(float), height, cudaMemcpyDeviceToHost);
Note this answer was added as a community wiki to get this question off the unanswered list