Calling initializer_list constructor via make_unique/make_shared

Quentin picture Quentin · Oct 15, 2014 · Viewed 12.3k times · Source

I'm trying to use std::make_unique to instanciate a class whose constructor is to receive an std::initializer_list. Here a minimal case :

#include <string>
#include <vector>
#include <initializer_list>
#include <memory>

struct Foo {
    Foo(std::initializer_list<std::string> strings) : strings(strings) {}

    std::vector<std::string> strings;
};

int main(int, char**) {

    auto ptr = std::make_unique<Foo>({"Hello", "World"});

    return 0;
}

You can see on Coliru that it doesn't build :

main.cpp:14:56: error: no matching function for call to 'make_unique(<brace-enclosed initializer list>)'
     auto ptr = std::make_unique<Foo>({"Hello", "World"});

So, is make_unique reportedly unable to use initializer_lists ? Is there a bug in GCC 4.9.1 ? Or did I overlook something ?

Answer

Kerrek SB picture Kerrek SB · Oct 15, 2014

std::make_unique is a function template which deduces the argument types which are passed to the object constructor. Unfortunately, braced lists are not deducible (with an exception for auto declarations), and so you cannot instantiate the function template when that missing parameter type.

You can either not use std::make_unique, but please don't go that route – you should avoid naked news as much as you can, for the children's sake. Or you can make the type deduction work by specifying the type:

  • std::make_unique<Foo>(std::initializer_list<std::string>({"Hello", "World"}))

  • std::make_unique<Foo, std::initializer_list<std::string>>({"Hello", "World"})

  • auto il = { "Hello"s, "World"s }; auto ptr = std::make_unique<Foo>(il);

The last option uses the special rule for auto declarations, which (as I hinted above) do in fact deduce an std::initializer_list.