std::move on a std::future stored in a pair

Question

I have a map:

    typedef std::map<std::string, std::pair<std::vector<SomeObject>, std::future<boost::optional<uint64_t>>>> t_map;
    t_map m;

When I insert in the map, I add a default constructed future to the pair:

...
std::future<boost::optional<uint64_t>> fut;
m.insert(t_map::value_type(mystring, std::make_pair(myvector, fut)));

Further on, I make a std::async call returning the future object which I then try to move to the pair:

auto record = m.find(somestring);
if (record != m.end())
{ 
    std::future<boost::optional<uint64_t>> f = std::async(fn, args);
    m->second.second(std::move(f));
}

However, I get the error:

error: no match for call to ‘(const std::future<boost::optional<long unsigned int> >) (std::remove_reference<std::future<boost::optional<long unsigned int> >&>::type)’
                         m->second.second(std::move(f));

I am not sure whats going wrong with this move semantics. Any advise?

UPDATE

I tried using operator= as Barry suggested, however I get the error no match to operator=. Why is the left hand operand being const qualified? I would have thought that the call will get resolved with the move assignment operator:

future& operator=(future&& __fut) noexcept

ERROR:

error: no match for ‘operator=’ (operand types are ‘const std::future<boost::optional<long unsigned int> >’ and ‘std::remove_reference<std::future<boost::optional<long unsigned int> >&>::type {aka std::future<boost::optional<long unsigned int> >}’)
                         m->second.second = std::move(f);
                                                 ^
note: candidates are:
In file included from test.cpp:20:0:
/.../linux/x86-64/release/opt/gcc-4.9.1/include/c++/4.9.1/future:687:15: note: std::future<_Res>& std::future<_Res>::operator=(const std::future<_Res>&) [with _Res = boost::optional<long unsigned int>] <near match>
       future& operator=(const future&) = delete;
               ^
note:   no known conversion for implicit ‘this’ parameter from ‘const std::future<boost::optional<long unsigned int> >*’ to ‘std::future<boost::optional<long unsigned int> >*’
/.../linux/x86-64/release/opt/gcc-4.9.1/include/c++/4.9.1/future:689:15: note: std::future<_Res>& std::future<_Res>::operator=(std::future<_Res>&&) [with _Res = boost::optional<long unsigned int>] <near match>
       future& operator=(future&& __fut) noexcept
               ^
note:   no known conversion for implicit ‘this’ parameter from ‘const std::future<boost::optional<long unsigned int> >*’ to ‘std::future<boost::optional<long unsigned int> >*’

Your most recent error seems to point to a `const` issue. It implies that `m` is `const`. Are you possibly accessing `m` as a member of a class in a method that is declared `const`? — md5i, May 20 '16 at 14:35
@md5i indeed, the const m was being returned by a const Getter function. — user2930006, May 21 '16 at 23:31

score 4 · Accepted Answer · answered May 20 '16 at 12:34

4

This:

m->second.second(std::move(f));

should be:

m->second.second = std::move(f);

As is, you're trying to call the future object. But future isn't callable, hence the error about "no match for call to ..."

answered May 20 '16 at 12:34

Barry

286,269
29
621
977

std::move on a std::future stored in a pair

1 Answers1

Linked