在构造函数初始化器列表中使用大括号了解奇怪的语法 [英] Understanding the weird syntax with curly braces in a constructor initializer list

问题描述

因此遇到此问题时，我只是在浏览库的源代码。

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)}
    {
    }

我不理解语法

m_font{..}

这是什么？它有什么作用。如果这真是愚蠢的问题，我感到抱歉。我不知道该对Google使用什么，所以在这里问。

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.

推荐答案

这在 cppreference ，但格式有点难以理解：

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

在复合语句的大括号之前，任何构造函数的函数定义的主体可能包括成员初始化器列表，其语法为冒号：，然后是一个或多个 member-initializers 的逗号分隔列表，每个列表都有以下语法

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

...

类或标识符大括号初始化列表（2）_{（自C ++ 11）}

...

2）初始化由命名的基或成员 class-or-identifier 使用列表初始化（如果列表为空，则变为值初始化；初始化聚集时，聚合初始化门）

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)

这是要说的是 X :: X（...）：some_member {some_expressions} {...} 导致 some_member 类成员从 some_expressions 。给定

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} {}
};

数据成员 y 将被初始化完全相同，将初始化局部变量 Y y {3}; 。

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

在您的情况下， std :: make_shared< sf :: Font>（字体）产生将用于初始化 m_font 类成员的值

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

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