省略号出现在模板函数的参数声明中

Question

This is the example from cppreference . I don't understand how the pattern get expanded.

template<typename ...Ts, int... N> void g(Ts (&...arr)[N]) {}
int n[1];
g<const char, int>("a", n); // Ts (&...arr)[N] expands to 
                            // const char (&)[2], int(&)[1]

Note: In the pattern Ts (&...arr)[N], the ellipsis is the innermost element, not the last element as in all other pack expansions.

Question 1: what is arr?

Question 2: n is a int array, does it match to int...N?

Question 3: How come it can expand to const char (&)[2], int(&)[1]

Answer 1

Whereas

template <typename ...Ts> void f(Ts&...arr);

is mostly equivalent to

template <typename T0, typename T1, .., typename TN>
void f(T0& arr0, T1& arr1, .., TN& arrN);

for any N .

In the same way,

template <typename ...Ts, int... Ns> void g(Ts (&...arr)[Ns]);

would be equivalent to

template <typename T0, typename T1, .., typename TN, int N0, int N1, .. int NN>
void g(T0 (&arr0)[N0], T1 (&arr1)[N1], .., TN (&arrN)[NN]);

and type T (&)[N] is a reference to C-array of size N with element of type T

int n[1]; is trivially of type int [1] .

"a" is of type const char[2] ( {'a', '\\0'} ).