Template argument deduction is done by comparing each function template parameter type (call it P) with the type of the corresponding argument of the call (call it A) as described below. If P is a dependent type, removing references and cv-qualifiers from P gives std::initializer_list<P'> or P'[N] for some P' and N, and the argument is an a non-empty initializer list (9.5.5 [dcl.init.list]), then deduction is performed instead for each element of the initializer list, taking P' as a function template parameter type and the initializer element as its argument , and in the P'[N] case, if N is a non-type template parameter, N is deduced from the length of the initializer list. Otherwise, an initializer list argument causes the parameter to be considered a non-deduced context (13.10.3.6 [temp.deduct.type]). [Example:
template<class T> void f(std::initializer_list<T>);
f({1,2,3}); // T deduced to int
f({1,"asdf"}); // error: T deduced to both int and const char*
template<class T> void g(T);
g({1,2,3}); // error: no argument deduced for T
template<class T, int N> void h(T const(&)[N]);
h({1,2,3}); // T deduced to int, N deduced to 3
template<class T> void j(T const(&)[3]);
j({42}); // T deduced to int, array bound not considered
struct Aggr { int i; int j; };
template<int N> void k(Aggr const(&)[N]);
k({1,2,3}); // error: deduction fails, no conversion from int to Aggr
k({{1},{2},{3}}); // OK, N deduced to 3
template<int M, int N> void m(int const(&)[M][N]);
m({{1,2},{3,4}}); // M and N both deduced to 2
template<class T, int N> void n(T const(&)[N], T);
n({{1},{2},{3}},Aggr()); // OK, T is Aggr, N is 3
—end example] For a function parameter pack...
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