Calling constructor with braces

Question

Simple question about C++11 syntaxis. There is a sample code (reduced one from source)

struct Wanderer
{
  explicit Wanderer(std::vector

        

Answer 1

It is just C++11 syntax. You can initialize objects calling their constructor with curly braces. You just have to bear in mind that if the type has an initializer_list constructor, that one takes precedence.

Answer 2

It is neither initializer list, nor uniform initialization. What's the thing is this?

Your premise is wrong. It is uniform initialization and, in Standardese terms, direct-brace-initialization.

Unless a constructor accepting an std::initializer_list is present, using braces for constructing objects is equivalent to using parentheses.

The advantage of using braces is that the syntax is immune to the Most Vexing Parse problem:

struct Y { };

struct X
{
    X(Y) { }
};

// ...

X x1(Y()); // MVP: Declares a function called x1 which returns
           // a value of type X and accepts a function that
           // takes no argument and returns a value of type Y.

X x2{Y()}; // OK, constructs an object of type X called x2 and
           // provides a default-constructed temporary object 
           // of type Y in input to X's constructor.