Emplacement of a vector with initializer list

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.}); 

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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.