java构造函数后面大括号,用大括号调用构造函数

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

struct Wanderer

{

explicit Wanderer(std::vector<:function>> & update_loop)

{

update_loop.emplace_back([this](float dt) { update(dt); });

}

void update(float dt);

};

int main()

{

std::vector<:function>> update_loop;

Wanderer wanderer{update_loop}; // why {} ???

}

I'd like to know, how it can be possible call constructor with curly brackets like Wanderer wanderer{update_loop}; It is neither initializer list, nor uniform initialization. What's the thing is this?

解决方案

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.