英文:

Why are curly braces {} NOT considered an operator in C++?

问题 {#heading}

我已查看官方操作符列表，但在任何操作符列表（无论是可重新定义还是不可重新定义的）中都没有看到{}出现。

据我所知，{}用于指定代码块，并且用于创建大括号初始化列表，但为什么在C++中不被视为操作符呢？ 英文:

I've checked official operator list, but didn't see {} appear in any list of operator (either redefinable or non-redefinable).

As far as I know, {} is used to specify a block of code, and also used to create a braced-init-list, but why is NOT considered an operator in C++?

答案1 {#1}

得分: 1

使用 {} 和诸如 +、!、() 等操作符的用法都属于表达式。具体来说，语言的语法包含以下规则（还有许多与表达式相关的规则）：

后缀表达式 :
后缀表达式 ( 表达式列表 ~可选~ )
简单类型说明符 花括号初始化列表

这里的第一条规则匹配使用 ()（调用操作符）的函数调用，例如 foo(0)。第二条规则匹配使用给定类型的列表初始化，例如 int{0}。

{} 和 () 在语法上非常相似，事实上，将花括号初始化列表视为一个操作符也并不远离事实。然而，有两个特征可以区分操作符：

1. 操作符通常是可组合的，这意味着你可以构建表达式，比如 x() + y()、(&x)() 和 x[0]()。相比之下，花括号初始化列表 只能应用于类型名称，用于初始化。{} 不能应用于另一个表达式。
2. 操作符通常是可以重载的（有一些例外）。列表初始化要么执行某种内置初始化，要么调用构造函数。它不能使用 operator{} 进行重载，这与操作符的预期行为不同。

英文:

Uses of {}, and uses of operators like +, !, (), etc. are expressions. Specifically, the language grammar contains the rule (among many others related to expressions):

> postfix-expression :<br>
> &nbsp;&nbsp;&nbsp;&nbsp;postfix-expression ( expression-list _opt )<br>
> &nbsp;&nbsp;&nbsp;&nbsp;simple-type-specifier &nbsp; braced-init-list <br>
> &nbsp;&nbsp;&nbsp;&nbsp;[...]

• [expr.post.general]

The first rule matches function calls with () (call operator) such as foo(0). The second rule matches list initialization with a given type, such as int{0}.

{} and () are quite similar gramatically, and it's not far from the truth to think of braced-init-list as an operator too. However, there are two features that distinguish operators:

1. Operators are usually composable, meaning that you can form expressions like x() + y(), (&x)(), and x[0](). By comparison, braced-init-lists can only be applied to the name of the type, to do initialization. {} cannot be applied to another expression.
2. Operators are (with some exceptions) overloadable. List initialization is either performing some built-in initialization, or it calls a constructor. It is not overloadable using operator{}, as one would expect from an operator.