Boost threading/mutexs, why does this work?
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by Flamewires
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Published on 2010-06-05T07:19:34Z
Indexed on
2010/06/05
7:22 UTC
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Code:
#include <iostream>
#include "stdafx.h"
#include <boost/thread.hpp>
#include <boost/thread/mutex.hpp>
using namespace std;
boost::mutex mut;
double results[10];
void doubler(int x) {
//boost::mutex::scoped_lock lck(mut);
results[x] = x*2;
}
int _tmain(int argc, _TCHAR* argv[])
{
boost::thread_group thds;
for (int x = 10; x>0; x--) {
boost::thread *Thread = new boost::thread(&doubler, x);
thds.add_thread(Thread);
}
thds.join_all();
for (int x = 0; x<10; x++) {
cout << results[x] << endl;
}
return 0;
}
Output:
0 2 4 6 8 10 12 14 16 18 Press any key to continue . . .
So...my question is why does this work(as far as i can tell, i ran it about 20 times), producing the above output, even with the locking commented out? I thought the general idea was:
in each thread: calculate 2*x copy results to CPU register(s) store calculation in correct part of array copy results back to main(shared) memory
I would think that under all but perfect conditions this would result in some part of the results array having 0 values. Is it only copying the required double of the array to a cpu register? Or is it just too short of a calculation to get preempted before it writes the result back to ram? Thanks.
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