Question

If multiple threads are simultaneously writing a single memory location.,there will be a race condition,right??
In my case same is happening..

Consider a module from ‘reduce.cl’

int i = get_global_id(0);
int n,j;

n = keyMobj[i];                       // this n is the key..It can be either 0 or 1.
for(j=0; j<2; j++)
      sumMobj[n*2+j] += dataMobj[i].dattr[j];        //summing operation.

Here, The memory locations
sumMobj===> […0…, ….1…] is accessed 4 threads simultaneously &
sumMobj===> [….3…, ….4…] is accessed 6 threads simultaneously..

Is there any way to still make it parallely,like using locking or semaphore? As this summing is a very big part in my algorithm…

Answer 1

I can give you some hint as I was also facing similar problem.

I can think of three different methods for achieving similar goal:

Consider a simple kernel, assuming you launched 4 (0-3) threads

_kernel void addition (int *p)
{
int i = get_local_id(0);
     p[4]+= p[i];
}

You want to add values p[0], p[1], p[2], p[3], p[4], and store the final sum in p[4]. right? i.e:

p[4]= p[0] + p[1] + p[2] + p[3] + p[4] 

Method -1 (no parallelism)

Assign this job to only 1 thread (no parallelism):

int i = get_local_id(0);
if (i==0)

{

p[4]+= p[i];

} 

Method-2 (with parallelism)

Express your problem as follows:

p[4]= p[0] + p[1] + p[2] + p[3] + p[4] + 0  

This is a reduction problem

So launch 3 threads: i=0 to i=2. In first iteration

 i=0 finds p[0] + p[1]
 i=1 finds p[2] + p[3]  
 i=2 finds p[4] + 0

Now you have three numbers, you apply the same logic as above and add these numbers (with suitable padding of 0 to make it in power of two)

Method -3 Atomic operations

If you still need to implement this atomically, you can use atomic_add():

  int fsfunc atomic_add (   volatile __global int *p ,int val)

Description

Read the 32-bit value (referred to as old) stored at location pointed
by p. Compute (old + val) and store result at location pointed by p.
The function returns old.

This is assuming the data is int type. Otherwise you can see the link as suggested above.