QtConcurrent::map segmentation fault

Question

When I have been trying to implement "parallel for" using QtConcurrent::map:

QFuture<void> parForAsync(size_t n, std::function<void (size_t)> Op)
{

    size_t nThreads =
            static_cast<size_t>(QThreadPool::globalInstance()->maxThreadCount());
    size_t nn = n/nThreads + 1;

    using Sequence = QVector<std::function<void()>>;
    Sequence vFuns;

    for(size_t i = 0; i < n; i+=nn)
    {
        size_t firstIdx = i,
                lastIdx = i + nn > n ? n : i + nn;

        vFuns.push_back([=]()->void
        {
            for(size_t i = firstIdx; i < lastIdx; ++i)
            {
                Op(i);
            }
        });
    }
    return QtConcurrent::map<Sequence>    //<-Segmentation fault!
            (vFuns, [](std::function<void()> f)
    {
        f();
    });
}

I've got segmentation fault in this place:

template<typename _Res, typename... _ArgTypes>
    function<_Res(_ArgTypes...)>::
    function(const function& __x)
    : _Function_base()
    {
      if (static_cast<bool>(__x))
    {
      __x._M_manager(_M_functor, __x._M_functor, __clone_functor); //<-Segmentation fault!
      _M_invoker = __x._M_invoker;
      _M_manager = __x._M_manager;
    }
    }

Why is this happening? It seems that std::function had passed checking. How can I make this code working?

Thanks in advance!

Answer 1

I cannot reproduce your case but I can give you some example to illustrate issue

QFuture<void> test ()
{   
    QVector<int> v; // LOCAL VARIABLE IN SCOPE OF test FUNCTION

    // preparing v vector

    QFuture<void> f = QtConcurrent::map(v,someFunction); // returns immediately
    return f;
}

[1] QtConcurrent::map takes v by reference NOT BY COPY.

[2] QtConcurrent::map returns immediately.

[3] So when test function ends, parallel operations started by map use v vector which was deleted because it is local variable in test function.

You can use waitForFinished for QFuture but then your function doesn't make sense because it blocks until parallel task ends.