how to use std::vector::emplace_back for vector<vector<int> >?

Question

  vector<vector<int> > res;
  res.emplace_back({1,2}); // change to res.push_back({1,2}); would work

This gives me error

main.cpp:61:25: error: no matching function for call to ‘std::vector<std::vector<int> >::emplace_back(<brace-enclosed initializer list>)’
main.cpp:61:25: note: candidate is:
In file included from /usr/include/c++/4.7/vector:70:0,
                 from /usr/include/c++/4.7/bits/random.h:34,
                 from /usr/include/c++/4.7/random:50,
                 from /usr/include/c++/4.7/bits/stl_algo.h:67,
                 from /usr/include/c++/4.7/algorithm:63,
                 from miscalgoc.hpp:1,
                 from main.cpp:1:
/usr/include/c++/4.7/bits/vector.tcc:92:7: note: void std::vector<_Tp, _Alloc>::emplace_back(_Args&& ...) [with _Args = {}; _Tp = std::vector<int>; _Alloc = std::allocator<std::vector<int> >]

How to make this work? Also, why an allocator is needed here?

Answer 1

The problem is that function template arguments doesn't deduce std::initializer_list from a braced-init-list (like { 1, 2 }).

Example:

#include <initializer_list>
#include <type_traits>


template<typename T>
void func(T arg) {
}


int main() {
    auto init_list = {1, 2};    // This works because of a special rule
    static_assert(std::is_same<decltype(init_list), std::initializer_list<int>>::value, "not same");

    func(std::initializer_list<int>{1, 2});     // Ok. Has explicit type.

    func({1, 2});   // This doesn't because there's no rule for function
                    //      template argument to deduce std::initializer_list
                    //      in this form.
}

Live example

std::vector::emplace_back() is a function template with its arguments being deduced. So passing it {1, 2} will not work because it couldn't deduce it. Putting an explicit type to it

res.emplace_back(std::initializer_list<int>{1,2});

would make it work.

Live example

Answer 2

@Mark's answer is pretty correct. Now let's consider a more practical case. After some proper operations, you've collected some data with vector<int>, and feel like pushing it into vector<vector<int>>:

std::vector<std::vector<int>> res;

for (int i = 0; i < 10000; ++i) {
    //
    // do something
    //
    std::vector<int> v(10000, 0);  // data acquired
    res.push_back(v);
}

It's not like assigning values you already know. Utilizing std::initializer_list is probably no longer a solution. In such cases, you may use std::move (along with either emplace_back or push_back is acceptable)

for (int i = 0; i < 10000; ++i) {
    std::vector<int> v(10000, 0);  // will become empty afterward
    res.emplace_back(std::move(v));  // can be replaced by 
                                     // res.push_back(std::move(v));
}

The performance is more or less improved. You can still be benefited from the concept of xvalue move-insertion, constructing objects by move-constructor rather than copying.

UPDATE

The reason that res.push_back(move(v)) works is because they overload the method std::vector::push_back(value_type&& val) after C++11. It is made to support rvalue reference deliberately.

Answer 3

Take a look at the documentation for vector::emplace_back. emplace_back tries to create a new element in your vector, by calling the constructor for the new element with the arguments passed in. So basially, when you call emplace_back({1,2}), it tries to pass {1,2} in to a constructor, but since res is a vector of vectors of ints, it's looking at vector constructors, none of which can take a brace-enclosed initializer list.

Also, take a look at the documentation for vector::push_back. When push_back is called, it creates a default object (in this case, a vector of ints) and copies the values into it. I would guess that the reason that push_back({1,2}) works is that the brace-enclosed initializer list creates a value-type, which push_back accepts.