Variadic template pack expansion

Viacheslav  Dronov picture Viacheslav Dronov · Sep 5, 2014 · Viewed 41.1k times · Source

I am trying to learn variadic templates and functions. I can't understand why this code doesn't compile:

template<typename T>
static void bar(T t) {}

template<typename... Args>
static void foo2(Args... args)
{
    (bar(args)...);
}

int main()
{
    foo2(1, 2, 3, "3");
    return 0;    
}

When I compile it fails with the error:

Error C3520: 'args': parameter pack must be expanded in this context

(in function foo2).

Answer

T.C. picture T.C. · Sep 5, 2014

One of the places where a pack expansion can occur is inside a braced-init-list. You can take advantage of this by putting the expansion inside the initializer list of a dummy array:

template<typename... Args>
static void foo2(Args &&... args)
{
    int dummy[] = { 0, ( (void) bar(std::forward<Args>(args)), 0) ... };
}

To explain the content of the initializer in more detail:

{ 0, ( (void) bar(std::forward<Args>(args)), 0) ... };
  |       |       |                        |     |
  |       |       |                        |     --- pack expand the whole thing 
  |       |       |                        |   
  |       |       --perfect forwarding     --- comma operator
  |       |
  |       -- cast to void to ensure that regardless of bar()'s return type
  |          the built-in comma operator is used rather than an overloaded one
  |
  ---ensure that the array has at least one element so that we don't try to make an
     illegal 0-length array when args is empty

Demo.

An important advantage of expanding in {} is that it guarantees left-to-right evaluation.


With C++17 fold expressions, you can just write

((void) bar(std::forward<Args>(args)), ...);