使用初始化列表放置向量

Question

i have a std::vector<std::vector<double>> and would like to add some elements at the end of it so this was my trial:

std::vector<std::vector<double> > vec;
vec.emplace_back({0,0});

but this does not compile whereas the following will do:

std::vector<double> vector({0,0});

Why can't emplace_back construct the element at this position? Or what am i doing wrong?

Thanks for your help.

Answer 1

The previous answer mentioned you could get the code to compile when you construct the vector in line and emplace that. That means, however, that you are calling the move-constructor on a temporary vector, which means you are not constructing the vector in-place, while that's the whole reason of using emplace_back rather than push_back .

Instead you should cast the initializer list to an initializer_list , like so:

#include <vector>
#include <initializer_list>

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

Answer 2

Template deduction cannot guess that your brace-enclosed initialization list should be a vector. You need to be explicit:

vec.emplace_back(std::vector<double>{0.,0.});

---

Note that this constructs a vector, and then moves it into the new element using std::vector 's move copy constructor. So in this particular case it has no advantage over push_back() . @TimKuipers 's answer shows a way to get around this issue.

Answer 3

There are really two issues here: the numeric type conversion and template argument deduction.

The std::vector<double> constructor is allowed to use the braced list (even of int s) for its std::vector<double>(std::initializer_list<double>) constructor. (Note, however, that std::initializer_list<int> does not implicitly convert to std::initializer_list<double> )

emplace_back() cannot construct the element from the brace expression because it is a template that uses perfect forwarding. The Standard forbids the compiler to deduce the type of {0,0} , and so std::vector<double>::emplace_back<std::initializer_list<double>>(std::initializer_list<double>) does not get compiled for emplace_back({}) .

Other answers point out that emplace_back can be compiled for an argument of type std::initializer_list<vector::value_type> , but will not deduce this type directly from a {} expression.

As an alternative to casting the argument to emplace_back , you could construct the argument first. As pointed out in Meyers' Item 30 (Effective Modern C++), auto is allowed to deduce the type of a brace expression to std::initializer_list<T> , and perfect forwarding is allowed to deduce the type of an object whose type was deduced by auto .

std::vector<std::vector<double> > vec;
auto double_list = {0., 0.}; // int_list is type std::initializer_list<int>
vec.emplace_back(double_list); // instantiates vec.emplace_back<std::initializer_list<double>&>

emplace_back adds an element to vec by calling std::vector<double>(std::forward<std::initializer_list<double>>(double_list)) , which triggers the std::vector<double>(std::initializer_list<double>) constructor.

Reference: Section 17.8.2.5 item 5.6 indicates that the this is a non-deduced context for the purposes of template argument deduction under 17.8.2.1 item 1.

Update: an earlier version of this answer erroneously implied that std::initializer_list<int> could be provided in place of std::initializer_list<double> .