what the difference between sizeof..(typename) and sizeof..(parameter) in variadic template c++

Question

Whats the difference between sizeof..(typename) and sizeof..(parameter) in a variadic template in c++?

In cpp primer 5th, I can't understand the difference:

#include <iostream>
using namespace std;

template <typename T, typename... Args>
void foo(const T &t, const Args &... rest)
{
    std::cout << "sizeof...(Args)=" << sizeof...(Args) << "  sizeof...(rest)=" << sizeof...(rest) << std::endl;
}

int main(int argc, const char **argv)
{
    int i = 0;
    double d = 3.14;
    string s = "how";

    foo(i, s, 42, d); /// sizeof...(Args)=3  sizeof...(rest)=3
    foo(s, 42, "hi"); /// sizeof...(Args)=2  sizeof...(rest)=2
    foo(d, s);        /// sizeof...(Args)=1  sizeof...(rest)=1
    foo("hi");        /// sizeof...(Args)=0  sizeof...(rest)=0
    foo(i, s, s, d);  /// sizeof...(Args)=3  sizeof...(rest)=3

    while (1)
        ;
    return 0;
}

Answer 1

sizeof...(rest) counts the number of arguments (not including the first). sizeof...(Args) counts the number of their types. There is no difference on the result.

Answer 2

The sizeof... operator gives the number of arguments in a parameter pack, and works for both template parameter packs (like Args ) and function parameter packs (like rest ). A function parameter pack always has the same count as the template parameter pack(s) used in declaring its type, for each specialization of the function template.