 # Simple Iterating Code -- HELP!

Hey everyone,
I have a code for a simple diffusion problem (with Q=5 iterations), which gives me an error message; at Line 15 (two times) and Line 18 (three times) it says, â€œerror: expression must have pointer-to-object typeâ€. Does anyone know what to do? (Again, I just want to run this diffusion equation for a given number of steps).
Thanks for looking!

# include <cuda.h>

global void incrementArrayOnDevice(float c, float q, int N)
{
int idx = blockIdx.x
int x;
int Q = 5;
for(x=0; x<Q; x++)
{
q[idx][x] = c[idx][x];
for(idx=1; idx<N-1; idx++)
{
c[idx][x] = 0.25
q[idx-1] + 0.5q[idx][x] + 0.25q[idx+1];
}
}
}
int main(void)
{
printf("\n");
float *a_h, *b_h;
float *a_d, q;
int i;
int N = 10;
size_t size = N
sizeof(float);
a_h = (float *)malloc(size);
b_h = (float *)malloc(size);
cudaMalloc((void **) &a_d, size);
cudaMalloc((void **) &q, size);
for (i=0; i<1; i++)
{
a_h[i] = 5;
}
for (i=1; i<N; i++)
{
a_h[i] = 0;
}
cudaMemcpy(a_d, a_h, sizeof(float)*N, cudaMemcpyHostToDevice);
int blockSize = 4;
int nBlocks = N/blockSize + (N%blockSize == 0?0:1);
incrementArrayOnDevice <<< nBlocks, blockSize >>> (a_d, q, N);
cudaMemcpy(b_h, a_d, sizeof(float)*N, cudaMemcpyDeviceToHost);

printf("\n");
for(i=0; i<N; i++)
{
printf("%f\n",b_h[i]);
}
free(a_h); free(b_h); cudaFree(a_d);
printf("\n");
}

Well, I don’t see an issue in the first function. Which compiler/system are you building with/on?

c[idx] = 0.25q[idx-1][x] + 0.5q[idx] + 0.25*q[idx+1];

The declaration of q is float *, but your trying to reference it as a 2 dimensional array. The compiler wouldn’t know how big each of the rows are to calculation the offset.

/Chris

Thanks, that puts me on the right track. Do you happen to know how I can specify the size of each row?

Compiler is “nvcc,” system is Mac OS X 10.5.

Simplest thing is to do your own index calculation:

So assuming a variable or define named W, this:

``````c[idx][x] = 0.25*q[idx-1][x] + 0.5*q[idx][x] + 0.25*q[idx+1][x];
``````

would become:

``````c[idx*W+x] = 0.25*q[(idx-1)*W+x] + 0.5*q[idx*W+x] + 0.25*q[(idx+1)*W+x];
``````

Since your looping you could get rid of the multiplication by just adding W to an index each time through the loop.

/Chris