std::async uses same thread and my code does not achieve parallelism. - c++

Question

I am using C++11 on Mac OS Xcode 4.3.2
std::async uses same thread and my code does not achieve parallelism. In sample code below I want to create 10 new threads. In each thread I want to calculate square root of input variable and set the result in promise. in main function I want to display the results calculated from threads. I am calling std::async with policy launch::async, So I expect it to create a new thread(10 times).
#include <mutex>
#include <future>
#include <thread>
#include <vector>
#include <cmath>
#include <iostream>
using namespace std;
mutex iomutex;
void foo(int i, promise<double> &&prms)
{
this_thread::sleep_for(chrono::seconds(2));
prms.set_value(sqrt(i));
{
lock_guard<mutex> lg(iomutex);
cout << endl << "thread index=> " << i << ", id=> "<< this_thread::get_id();
}
}
int main()
{
{
lock_guard<mutex> lg(iomutex);
cout << endl << "main thread id=>"<< this_thread::get_id();
}
vector<future<double>> futureVec;
vector<promise<double>> prmsVec;
for (int i = 0; i < 10; ++i) {
promise<double> prms;
future<double> ftr = prms.get_future();
futureVec.push_back(move(ftr));
prmsVec.push_back(move(prms));
async(launch::async, foo, i, move(prmsVec[i]));
}
for (auto iter = futureVec.begin(); iter != futureVec.end(); ++iter) {
cout << endl << iter->get();
}
cout << endl << "done";
return 0;
}
However if I use std::thread, then I can achieve parallelism.
#include <mutex>
#include <future>
#include <thread>
#include <vector>
#include <cmath>
#include <iostream>
using namespace std;
mutex iomutex;
void foo(int i, promise<double> &&prms)
{
this_thread::sleep_for(chrono::seconds(2));
prms.set_value(sqrt(i));
{
lock_guard<mutex> lg(iomutex);
cout << endl << "thread index=> " << i << ", id=> "<< this_thread::get_id();
}
}
int main()
{
{
lock_guard<mutex> lg(iomutex);
cout << endl << "main thread id=>"<< this_thread::get_id();
}
vector<future<double>> futureVec;
vector<promise<double>> prmsVec;
vector<thread> thrdVec;
for (int i = 0; i < 10; ++i) {
promise<double> prms;
future<double> ftr = prms.get_future();
futureVec.push_back(move(ftr));
prmsVec.push_back(move(prms));
thread th(foo, i, move(prmsVec[i]));
thrdVec.push_back(move(th));
}
for (auto iter = futureVec.begin(); iter != futureVec.end(); ++iter) {
cout << endl << iter->get();
}
for (int i = 0; i < 10; ++i) {
thrdVec[i].join();
}
cout << endl << "done";
return 0;
}

Answer 1

async(launch::async, foo, i, move(prmsVec[i]));
This line returns a future but because you do not assign it to anything the future's destructor runs at the end of the statement, which blocks and waits for the result by calling std::future::wait()
Why are you manually calling std::async with a promise, when it returns a future anyway? The point of async is that you don't need to manually use a promise, that's done internally for you.
Rewrite your foo() to return double then call it with async
#include <mutex>
#include <future>
#include <thread>
#include <vector>
#include <cmath>
#include <iostream>
using namespace std;
mutex iomutex;
double foo(int i)
{
this_thread::sleep_for(chrono::seconds(2));
lock_guard<mutex> lg(iomutex);
cout << "\nthread index=> " << i << ", id=> "<< this_thread::get_id();
return sqrt(i);
}
int main()
{
cout << "\nmain thread id=>" << this_thread::get_id();
vector<future<double>> futureVec;
for (int i = 0; i < 10; ++i)
futureVec.push_back(async(launch::async, foo, i));
for (auto& fut : futureVec)
{
auto x = fut.get();
lock_guard<mutex> lg(iomutex);
cout << endl << x;
}
cout << "\ndone\n";
}