Returning std:vector with std::async c++11

Question

I have a problem returning std::vector using std::future and std::async. Why does

#include <iostream>
#include <vector>
#include <functional>
#include <future>

using namespace std;

int main(int argc, char** argv) {

  int A = 10;
  vector<future<int>> sumarray(A);

  for(int i = 0; i < A; i++) 
    sumarray[i] = async( launch::async, [i] { return i; } );

  for(int i = 0; i < A; i++) 
    cout << sumarray[i].get() << " "; 
  cout << endl;

}

compilled with g++ -std=c++11 -pthread work as expected printing

0 1 2 3 4 5 6 7 8 9

but

#include <iostream>
#include <vector>
#include <functional>
#include <future>

using namespace std;

int main(int argc, char** argv) {

  int A = 10;
  int B = 2;
  vector<future<vector<int>>> sumarray(A);

  for(int i = 0; i < A; i++){
    sumarray[i] = async( launch::async, [i,B] { 
        vector<int> v(B);
        for(int j = 0; j < B; j++) v[j] = (j+1)*i;
        return v;
      }); 
  }

  for(int j = 0; j < B; j++)
    for(int i = 0; i < A; i++) 
      cout << sumarray[i].get()[j] << " "; 
  cout << endl;

}

compiled in the same way throw

terminate called after throwing an instance of 'std::future_error'
  what():  No associated state

Is there something wrong with the way I am returning the vector in the lambda function with std::async?

Answer 1

You are calling get() multiple times on the elements of sumarray. According to the documentation:

valid() is false after a call to this method.

This means that you can't call get() more than once as your inner loop does.