Consider the following code:
#include <iostream>
#include <vector>
#include <thread>
#include <mutex>
#include <future>
std::mutex mutex;
int generate()
{
static int id = 0;
std::lock_guard<std::mutex> lock(mutex);
id++;
std::cout << id << '\n';
return id;
}
int main()
{
std::vector<std::future<int>> vec;
std::vector<int> result;
for(int i = 0; i < 10; ++i)
{
vec.push_back(std::async(std::launch::async,generate));
}
for(int i = 0; i < 10; ++i)
{
result.push_back(vec[i].get());
}
std::cout << "\n result:";
for(const auto res : result)
{
std::cout << res << " ";
}
}
Which produces following output:
1
2
3
4
5
6
7
8
9
10
result:1 2 3 6 4 5 7 8 9 10
As you can see, the output inside generate()
function has correct (from 1 to 10) order. But when iterating over the result vector I'm getting different order each run. Why is this happening and how can I synchronize such code?
What you're seeing in the final output is the order in which the threads produced their results.
Consider these three threads as a simplified version:
+- -+ +- -+ +- -+
Thread | 1 | | 2 | | 3 |
+- -+ +- -+ +- -+
Result | | | | | |
+- -+ +- -+ +- -+
Now, no matter which order these execute and acquire the mutex in, the output will be
1 2 3
because id
is static
, and the first "locker" increments it and prints "1", then the second increments it and prints "2", and so on.
Let's assume that Thread 3 gets the mutex first, then 1, then 2.
That leads to this situation:
+- -+ +- -+ +- -+
Thread | 1 | | 2 | | 3 |
+- -+ +- -+ +- -+
Result | 2 | | 3 | | 1 |
+- -+ +- -+ +- -+
And when you print the results in "vector order", you will see
2 3 1
If you want threads to execute in some predetermined order, you need to synchronize them further.