I find one site that says it has 16 shaders, that would mean 2 multiprocessors if I am not mistaken. If you copy the data in the GPU sheet of G84 to the right and change the name & amount of multiproc to 2, you will probably get valid output.
Note that the occupancy is the same for all versions, it is just the output of Maximum Simultaneous Blocks per GPU, that differs between the versions. And that is not really very important, occupancy is the most important output of the sheet.
Why is it that the value displayed by nNum is 19? What I am trying to do is add the indices of the threads to nNum so the value is supposed to be 0+1+2+3…+19. It turns out that the value displayed is the addition of the last index of the thread to nNum.
This is my code.
main.cpp:
#include <stdio.h>
#include <stdlib.h>
#define NUM 20
extern “C” void test (int* nNum, int nSize);
int main(int argc, char *argv)
{
int temp = 0;
test(&temp, NUM);
printf (“%d\n”, temp);
}
threadstest.cu:
global void compute_testd(int* nNum, int nSize)
{
unsigned int index = threadIdx.x + blockIdx.x * blockDim.x;
if (index < nSize)
{
*nNum += index;
}
}
extern “C” void test (int* nNum, int nSize)
{
int* nNumd;
cudaMalloc((void**)&nNumd, sizeof(int));
cudaMemcpy(nNumd, nNum, sizeof(int), cudaMemcpyHostToDevice);
compute_testd<<< ceil((float)nSize/256.0f), 256>>> (nNumd, nSize);
cudaMemcpy(nNum, nNumd, sizeof(int), cudaMemcpyDeviceToHost);
cudaFree(nNumd);
}
And is there a way for us to put this loop into CUDA:
for (int i = 0; i < 20; i++)
{
for (int j = 0; j < 20; j++)
{
array[i][j] = 3;
}
}