parallelize a code?can't make it work right

Accelerated Computing CUDA CUDA Programming and Performance
1

I am trying to parallelize a code but I am not getting the right results.
The code creates data which creates an image.

The c code is like:

J = 0;
Constants = 0;
	for ( RowIdx = 0; RowIdx < Rows; RowIdx++ )
	{

		
		*(theRe + J) = 0.0;
		*(theIm + J) = 1.0;

		++Constants;

		++J;
		P = J + 1;

for ( ColIdx = 1; ColIdx < Cols; ColIdx++ )
   *(theRe + J) =   *(thePh + J) * ....
    ++J;

} //ColIdx

....

} //RowIdx

Since we do not use at all the RowIdx or ColIdx ,but we use “J”, I tried:

J = threadIdx.y + blockDim.y * blockIdx.y;
Constants = 0;
	for ( RowIdx = 0; RowIdx < Rows; RowIdx++ )
	{

		
		*(theRe + J) = 0.0;
		*(theIm + J) = 1.0;

		++Constants;

		J+= gridDim.y * blockDim.y;
		P = J + 1;

for ( ColIdx = 1; ColIdx < Cols; ColIdx++ )

but I am not getting the right results.

Also , I do not understand if I must use :

RowIdx = threadIdx.y + blockDim.y * blockIdx.y;
for ( ; RowIdx < Rows; RowIdx  += gridDim.y * blockDim.y)

(and the same for column)

Because I thought that we aren’t using anywhere the RowIdx or ColIdx ,it isn’t required to express them like threads.

2

Try illiminating both for loops, these should no longer be necessary.

Every element on an index will be processed by a cuda core/kernel.

3

The problem is that these 2 loops are nested.And they access same variables.

If I leave them as it is,its is wrong because

J+= gridDim.y * blockDim.y;

will be in a loop also.

But how can I remove them? (updating the code)

4

Wouldn’t it be best to use 2D grid of blocks in which every thread do calculations for single pixel (or a tile of them) of the image? Could You elaborate what exactly You want to do, if it is not a secret? :)

Cheers,
MK

5

I tried to use

RowIdx = threadIdx.y + blockDim.y * blockIdx.y;
ColIdx = threadIdx.x + blockDim.x * blockIdx.x;

but then I don’t know how to handle “J” ans “P”. (if you look at c code).
J and P are neighboor values.

It is a little bit of secret indeed! :)

The code creates binary data and I can read them as image.
The code is repeating (the c code I have ).

Thanks