英文:
Why are curly braces {} NOT considered an operator in C++?
问题
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
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）：
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：
- 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。 - 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>
> postfix-expression ( expression-list<sub>opt</sub> )<br>
> simple-type-specifier braced-init-list<br>
> [...]
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:
- Operators are usually composable, meaning that you can form expressions like
x() + y(),(&x)(), andx[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. - 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.
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