如何将可变参数列表传递给接受可变参数的函数？

Question

I have a template parameter pack that I would like to pass into a function that takes in a variable number of parameters. Since I don't know how many parameters I actually have, how can I pass in an unknown amount of parameters' data members into the function?

template <typename BarType...>
Class Test
{
   explicit Test(BarType ...bars)
   {
   // Calling this function with parameter pack instead of this way 
   // (which won't even work with above parameter pack)
   FooFunction(
   [](BarType &bar1){ return bar1.bar_member; },
   [](BarType &bar2){ return bar2.bar_member; },
   [](BarType &bar3){ return bar3.bar_member; })
   }

}

How can I accomplish the above FooFunction without manually listing out all three parameters (making it more dynamic)? Since there might be more parameters bar4 , bar5 etc. So just manually listing them won't work.

Answer 1

It looks like you might want:

template <typename BarType...>
Class Test
{
   explicit Test(BarType ...bars)
   {
       FooFunction(bars.bar_member ...);
   }
};

The most common way of using a parameter pack is to expand it, by putting a representation of what should be in each element of a list, followed by the ellipsis token ... .

So if we take the specialization Test<B1, B2, B3> , it has a constructor like explicit Test(B1 b1, B2 b2, B3 b3); except that the names b1 , b2 , and b3 are just made up for explanation. In that function body, some possible things involving pack expansions and what they would instantiate as:

std::make_tuple(bars...) -> std::make_tuple(b1, b2, b3)
std::tuple<BarType...>   -> std::tuple<B1, B2, B3>
f(g(bars) ...)           -> f(g(b1), g(b2), g(b3))
h<BarType>(bars)...      -> h<B1>(b1), h<B2>(b2), h<B3>(b3)

The last one expands both the template parameter pack BarType and the function parameter pack bars together, matching up the corresponding pack elements.