Constructors taking initializer lists

Question

I get the idea behind uniform initialization with curly braces. But why is that using this syntax on types that have a constructor that takes an initializer list, calls that specific constructor, even though the arguments are only wrapped in a single pair of curly braces, i.e.

int main(int argc, const char ** argv)
{
    vector<int> vs0{3};

    for(int v : vs0)
    {
        cout << v << ' ';
    }

    cout << '\n';

    vector<int> vs1(3);

    for(int v : vs1)
    {
        cout << v << ' ';
    }
}

/*
    Output
    3
    0 0 0
*/

Why does vs0 get constructed with the initializer list constructor? Shouldn't it be

vector<int> v2{{3}};

for that? It's kind of confusing, especially if you're not aware that a class has a constructor that takes an initializer list.

Answer 1

If the class has a constructor taking std::initializer_list, then it'll be preferred when being passed a braced-init-list in list initliazation.

• Otherwise, the constructors of T are considered, in two phases:

  • All constructors that take std::initializer_list as the only argument, or as the first argument if the remaining arguments have default values, are examined, and matched by overload resolution against a single argument of type std::initializer_list

  • If the previous stage does not produce a match, all constructors of T participate in overload resolution against the set of arguments that consists of the elements of the braced-init-list, with the restriction that only non-narrowing conversions are allowed. If this stage produces an explicit constructor as the best match for a copy-list-initialization, compilation fails (note, in simple copy-initialization, explicit constructors are not considered at all).

{3} is a braced-init-list, then the vector will be initialized as containing 1 element with value 3. The behavior is consistent with passing {1, 2}, {1, 2, 3} and so on.