emplace_back not working with std::vector>

Question

I am trying to do emplace_back into a std::vector>, but could not find the right syntax to do it.

#i         


        

Answer 1

emplace_back does forward all arguments to a matching constructor of the member type. Now, std::map has a initializer-list constructor, but it expects a list of std::pair, i.e. std::pair. push_back is not a template, so it just expects one type and thus performs the conversion in place. That is, no type-deduction occurs here.

You would need to explicitly state that you want to have a std::pair; the following should work:

#include

For the same reason, this does not compile:

    v.emplace_back({std::pair(1,2),
                    std::pair(3,4)});

This is because, though a brace-enclosed list may yield an initializer-list, it doesn't have to. It can also be a constructor call or something like that. So, writing

auto l = {std::pair(1,2),
          std::pair(3,4)};

yields an initializer list for l, but the expression itself might be used in another way:

std::pair, std::pair> p =
          {std::pair(1,2),
          std::pair(3,4)}

This whole stuff gets a bit messy.

Basically, if you have an brace-enclosed-list, it may yield an initializer list or call a matching constructor. There are cases where the compiler is not able to determine which types are needed; emplace_back is one of them (because of forwarding). In other cases it does work, because all types are defined in the expression. E.g.:

#include 
#include 

int main() 
{
    std::vector> v = 
         {{1,2},{3,4},{5,6}};
    return 0;
}

Now the reason it doesn't work is that no type can be deduced. I.e. emplace_back tries to deduce the name of the input types, but this is not possible, since a brace-enclosed-list has several types it can describe. Hence there is not a matching function call.