romants
romants

Reputation: 3648

OpenMP implementation of reduction

I need to implement reduction operation (for each thread the value should be stored in different array entry). However, it runs slower for more threads. Any suggestions?

double local_sum[16];.
//Initializations....
#pragma omp parallel for shared(h,n,a) private(x, thread_id) 
for (i = 1; i < n; i++) {
    thread_id = omp_get_thread_num();
    x = a  + i* h;
    local_sum[thread_id] += f(x);
}

Upvotes: 0

Views: 1297

Answers (2)

Hristo Iliev
Hristo Iliev

Reputation: 74385

You are experiencing the effects of false sharing. On x86 a single cache line is 64 bytes long and therefore holds 64 / sizeof(double) = 8 array elements. When one thread updates its element, the core that it runs on uses the cache coherency protocol to invalidate the same cache line in all other cores. When another thread updates its element, instead or operating directly on the cache, its core has to reload the cache line from an upper-level data cache or from the main memory. This significantly slows down the program execution.

The simplest solution is to insert padding and thus spread array elements accessed by different threads into distinct cache lines. On x86 that would be 7 double elements. Therefore your code should look like:

double local_sum[8*16];
//Initializations....
#pragma omp parallel for shared(h,n,a) private(x, thread_id) 
for (i = 1; i < n; i++) {
    thread_id = omp_get_thread_num();
    x = a  + i* h;
    local_sum[8*thread_id] += f(x);

}

Don't forget to take only each 8th element when summing the array at the end (or initialise all array elements to zero).

Upvotes: 5

pavel.medyankin
pavel.medyankin

Reputation: 1

Did you try to use reduction?

double global_sum = 0.0;
#pragma omp parallel for shared(h,n,a) reduction(+:global_sum) 
for (i = 1; i < n; i++) {
    global_sum += f(a  + i* h);
}

Howerver there may be a lot of other reasons why it runs slow. For example you should not create 16 threads if you have only 2 CPU cores and so on.

Upvotes: -2

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