Question

Somehow I don’t get how variadic template parameter packs are expanded. What’s wrong with thie following code?

#include <iostream>

template <typename T>
struct print_one
{
    static void run(const T& t)
    {
        std::cout << t << ' ';
    }
};

template<typename... Args>
void print_all(Args&&... args)
{
    // the next line doesn't compile:
    print_one<Args>::run(std::forward<Args>(args))...;
}

int main()
{
    print_all(1.23, "foo");
}

Clang says, Expression contains unexpanded parameter packs 'Args' and 'args'. Why?

Answer 1

The ... has to go inside the function call parentheses:

print_one<Args>::run(std::forward<Args>(args)...);

Obviously, that won’t work for your function that takes only a single argument, so you need to find a way to expand the calls into a function call or other allowed construct:

// constructing a dummy array via uniform initialization
// the extra 0 at the start is to make it work when the pack is empty
int dummy[]{0, (print_one<Args>::run(std::forward<Args>(args)), 0)...};

// or, if your compiler doesn't support uniform initialization
int dummy[] = {0, (print_one<Args>::run(std::forward<Args>(args)), 0)...};

// or, calling a dummy function
template<typename... Args> void dummy(Args...) {}
dummy((print_one<Args>::run(std::forward<Args>(args)), 0)...);

// or, constructing a temporary dummy object
struct dummy { dummy(std::initializer_list<int>) {} };
dummy{(print_one<Args>::run(std::forward<Args>(args)), 0)...};

// or, constructing a temporary initializer list
std::initializer_list<int>{(print_one<Args>::run(std::forward<Args>(args)), 0)...};

Note the use of the comma operator to turn the void return of print_one into a value suitable to place in an argument list or initializer expression.

The initializer-list forms are preferred to the function call forms, as they are (supposed to be) ordered LTR which function call arguments are not.

The forms where a parameter pack expansion can occur are covered by 14.5.3 [temp.variadic]:

4 – […] Pack expansions can occur in the following contexts:

• […]

Your original code is illegal because although textually it might appear that it should produce a statement consisting of a number of comma-operator expressions, that is not a context allowed by 14.5.3:4.