What is the precedence of the meta-operator ...?

Question

What is the precedence of the meta-operator ... whose job is to unpack template type parameter packs? I imagine it's pretty low, but how low is it? The C++ standard says:

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

Anyone up for the challenge? Of course, ... does not appear in C++03 operator precedence tables.

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Okay, if ... is not an operator, what exactly determines that std::forward<Args>(args)... applies to the the entire sequence std::forward<Args>(args) and not just (args), for example?

Answer 1

It doesn't seem to be an operator. From N3092 (sorry I don't have a more recent draft handy)

[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]