c++ - 元运算符的优先级是什么...？

Question

元运算符...的优先级是多少？谁的工作是解压模板类型参数包？我想它很低，但它有多低？ C++ 标准说:

The precedence of operators is not directly specified, but it can be derived from the syntax.

有人愿意接受挑战吗？当然，...没有出现在 C++03 运算符优先级表中。

---

好的，如果...不是运算符，究竟是什么决定了 std::forward<Args>(args)...适用于整个序列 std::forward<Args>(args)而不仅仅是 (args) , 例如?

Answer 1

好像不是运算符。来自 N3092(抱歉，我手边没有更新的草稿)

[14.5.3] 4/ A pack expansion is a sequence of tokens that names one or more parameter packs, followed by an ellipsis. The sequence of tokens is called the pattern of the expansion; its syntax depends on the context in which the expansion occurs. Pack expansions can occur in the following contexts:

• In an initializer-list (8.5); the pattern is an initializer-clause.
• In a base-specifier-list (10); the pattern is a base-specifier.
• In a mem-initializer-list (12.6.2); the pattern is a mem-initializer.
• In a template-argument-list (14.3); the pattern is a template-argument.
• In a dynamic-exception-specification (15.4); the pattern is a type-id.
• In an attribute-list (7.6.1); the pattern is an attribute.

• In a capture-list (5.1.2); the pattern is a capture. [Example:

  template<class ... Types> void f(Types ... rest);
  template<class ... Types> void g(Types ... rest) {
      f(&rest ...); // “&rest ...” is a pack expansion; “&rest” is its pattern
  }
  

  — end example]