First, sorry for the delay !
Then, I compiled a small cuda program running the same code on the cpu and the gpu to compare results.
#include <stdio.h>
#include <cuda.h>
#define SBA(sb, v) (*(unsigned int*)((char*)(sb)+(v)))
__host__ void mykernel_cpu()
{
int i, tab[5];
for(i=0; i<5; i++)
tab[i] = 15795 * i;
for(i=0; i<5; i++)
printf("%d = %ld\n", i, tab[i]);
printf("=> %ld\n", SBA(tab, 1 & 0xffff));
printf("=> %ld\n", SBA(tab, 3 & 0xffff));
}
__global__ void mykernel_gpu()
{
int i, tab[5];
if( threadIdx.x == 0 )
{
for(i=0; i<5; i++)
tab[i] = 15795 * i;
//__syncthreads();
for(i=0; i<5; i++)
printf("%d = %ld\n", i, tab[i]);
printf("=> %ld\n", SBA(tab, 1 & 0xffff));
printf("=> %ld\n", SBA(tab, 3 & 0xffff));
}
}int main( void )
{
printf("\n-----------------------\n");
printf("\tCPU RUN");
printf("\n-----------------------\n");
mykernel_cpu();
printf("\n-----------------------\n");
printf("\tGPU RUN");
printf("\n-----------------------\n");
dim3 dimGrid(5,5,1);
dim3 dimBlock(2,2,2);
mykernel_gpu <<< dimGrid, dimBlock >>> ();
printf("\n");
return 1;
}
Results from CPU and GPU are the same. So as you mentionned it, it’s not an endian issue.
And as you can see, this is nor a sync issue.
The point is i’m porting a big CPU code to GPU. I think there is another tricky issue in my portage. I will investigate in depth and come back if I need. However, thank you for your time & help.
