What are the 6 dots in template parameter packs? [duplicate]

Question

This question already has an answer here:

• What is the meaning of “… …” token? i.e. double ellipsis operator on parameter pack 2 answers

While looking at this question I found myself in the cpp reference site where I noticed a strange and new to me syntax :

template<class Ret, class... Args>
struct is_function<Ret(Args......)volatile &&> : std::true_type {};

Yep, 6 dots ! Initially I thought this was a typo, but after checking the libstdc++ source again there it was eg at line 444 :

template<typename _Res, typename... _ArgTypes>
struct is_function<_Res(_ArgTypes......) volatile &&> : public true_type { };

Is this a valid syntax ? Dot dot dot, are used to pack and unpack parameter packs ? What do 6 dots do ?

Answer 1

In this case, the two are for different purposes. The first is for parameter pack expansion and the second is for variable argument lists. That particular declaration is to handle functions which take some regular parameters plus a variable argument list.

The difference is between run-time and compile-time variability. A function which takes a variable number of arguments at run-time is special. It is a single function which can handle a variable number of arguments from the caller:

void f(int x,...) // equivalent to void f(int x ...)
{
    // Do some run-time logic here to determine what to
    // do with parameters after x.
}

This is distinct from the notion that we want to be able to have a template which uses a variety of functions with various parameters which are known at compile time. For example, we could define a function template which takes a pointer to a function and allow the number and types of the arguments to vary:

template <typename... Args>
void g(void (*function_ptr)(Args...)) 
{ 
    // We can do logic here to call function_ptr with the proper
    // number of arguments.
}

Given these functions:

void f1(int);
void f2(int,float);

You can call g with any of them:

g(f1); // fine
g(f2); // also fine

However

g(f); // error

The compiler wouldn't know what to use for the Args parameter pack in g.

Answer 2

Why does libstdc++ use ... ... in it's implementation of is_function? If we check out the cppreference section for std::is_function it gives a sample implementation and says for the first ... ... case:

// specialization for variadic functions such as std::printf
template<class Ret, class... Args>
struct is_function<Ret(Args......)> : std::true_type {};

so we need the second set of ... to match a variadic function like printf:

           Comma optional as per 8.3.5 [dcl.fct] 
           |
           v
Ret(Args... ...)
        ^   ^
        |   |
        Match a function with a variable number of arguments
            |
            and the function is a variadic function

Note, we have functions like fprintf that two arguments before the variadic terms and we need to match those as well. Indeed if we use that implementation and attempt to match printf without the ... ... specialization then it fails see it live.

This corner of the language is covered in this post C++11's six dots:

I was mucking around the other day and discovered this nice little oddity:

template <typename... Args>
void foo(Args......);

As it turns out, ...... can be totally valid C++11. This is what happens when backward compatibility mixes with new hotness.

// These are all equivalent.

template <typename... Args> void foo1(Args......);
template <typename... Args> void foo2(Args... ...);
template <typename... Args> void foo3(Args..., ...);

Hopefully the last one shows what is happening here. [...]

Why is this valild? We can see that , ... is synonymous with ... from the draft C++11 standard section 8.3.5 [dcl.fct] which has the following grammar:

parameter-declaration-clause:
  parameter-declaration-listopt...opt
  parameter-declaration-list , ...

and says:

[...] Where syntactically correct and where “...” is not part of an abstract-declarator, “, ...” is synonymous with “...”. [...]