Why OpenMP lastprivate produces wrong result? - c

Question

I am trying to run following code to understand the functionality of OpenMP lastprivate construct. As per the definition of lastprivate, if I declare a variable lastprivate, it is private to every thread and the value of the thread executing the last iteration of a parallel loop in sequential order is copied to the variable outside of the region.
Here is the code:
int main(void)
{
omp_set_num_threads(5);
int i;
int k =3;
#pragma omp parallel private(i)
{
#pragma omp for lastprivate(k)
for(i=0; i< 5; i++ )
{
int iam = omp_get_thread_num();
k = iam;
printf("k=%d, iam=%d\t",k, iam);
}
}
printf("\n k = %d", k);
}
It produces output something like this:
k=0, iam=0 k=4, iam=4 k=3, iam=3 k=2, iam=2 k=1, iam=1
k = 4
When we have a team of threads working in a 'for', we cannot really guarantee which thread executes last. So, accordingly, the value of last thread should reflect in global 'k'. However, no matter how many times I run the code , the value of 'k' globally (i.e. after the parallel section is over) remains 4.
From the printed values too, we can see that thread 1 executed last. Even if we assume that the prints are not reliable to get the exact running sequence of threads, it seems far from obvious that the thread 4 always runs last, thereby reflecting its value in 'k'.
I would appreciate the help regarding this problem. Thanks.

Answer 1

To be sure which thread executes last, you should print the value of the iteration index (and not replicate the thread id):
#include<stdio.h>
#include<omp.h>
int main() {
int kk;
#pragma omp parallel
{
#pragma omp for schedule(runtime) lastprivate(kk)
for(int ii=0; ii < 1000; ii++ ) {
kk = omp_get_thread_num();
printf("ii = %d, kk = %d\n",ii,kk);
}
}
printf("kk = %d\n", kk);
return 0;
}
If you run this program you will notice that the thread that executes iteration 999 sets the value of kk.
Regarding this sentence (emphasis mine):
When we have a team of threads working in a 'for', we cannot really guarantee which thread executes last.
What you say is generally true, but with one exception (section 2.5. of the OpenMP 3.1 standard):
Different loop regions with the same schedule and iteration count,
even if they occur in the same parallel region, can distribute
iterations among threads differently. The only exception is for the
static schedule...
Now, as you didn't specify any schedule, the following rule holds:
If the loop directive does not have a schedule clause then the current
value of the def-sched-var ICV determines the schedule
If def-sched-var determines a schedule(static) (as I have experienced is many times the case) then the final print of your program will be always k = 4

Answer 2

You are confusing two different ideas of "last".
The standard says "when a lastprivate clause appears on the directive that identifies a worksharing construct, the value of each new list item from the sequentially last iteration of the associated loops"
That says nothing about the order in which things execute, whereas you are assuming that "last" means the temporally last thread to execute.
So if you have static loop scheduling that guarantees that the highest numbered thread will execute the last loop iteration, therefore the value saved will always be that from the highest numbered thread, and it has nothing to do with the specific (random) order in which the threads happened to execute.