在构造函数初始化程序列表中使用大括号了解奇怪的语法

Question

So I was just browsing the source code of a library when I encountered this.

Font::Font(const sf::Font& font) :
        m_font{std::make_shared<sf::Font>(font)}
    {
    }

I don't understand the syntax

m_font{..}

What is it? What does it do. I am sorry if this is really stupid question. I don't know what to Google, so asking here.

Answer 1

This is described on cppreference , but in a somewhat hard to read format:

The body of a function definition of any constructor, before the opening brace of the compound statement, may include the member initializer list , whose syntax is the colon character : , followed by the comma-separated list of one or more member-initializers , each of which has the following syntax

...

class-or-identifier brace-init-list (2) _{(since C++11)}

...

2) Initializes the base or member named by class-or-identifier using list-initialization (which becomes value-initialization if the list is empty and aggregate-initialization when initializing an aggregate)

What this is trying to say is that X::X(...) : some_member{some_expressions} { ... } causes the some_member class member to be initialised from some_expressions . Given

struct X {
    Y y;
    X() : y{3} {}
};

the data member y will be initialised the exact same way a local variable Y y{3}; would be initialised.

In your case, std::make_shared<sf::Font>(font) produces the value that will be used to initialise the m_font class member.

Answer 2

That is a list initialization aka brace initializer list. More specifically, in this case it's a direct-list initialization.

Basically the m_font variable is initialized with the value that's given in the curly braces, in this case it's initialized to a shared_ptr for the font object that's given to the constructor.

Answer 3

Font类具有一个名为m_font _ std::shared_ptr<sf::Font>类型的m_font成员，因此在Font类的构造函数中，该成员正在使用共享的font指针进行初始化。