c++ - "typename qualified-id" referring to a type in a non-type parameter-declaration -


14.1 [temp.param], paragraph 2

... typename followed unqualified-id names template type parameter. typename followed qualified-id denotes type in non-type parameter-declaration...

i'm bit confused meaning of bold text. specifically, typename can occur in 2 different contexts (type-specifier or template-parameter), 1 refer to?

  • for former case, considered:

    struct {     struct x { };     int x; }; struct b {     struct x { }; }; template<class t> void f(t t) {     typename t::x x;    // t can or b }

    however, neither a::x nor b::x non-type parameter-declaration (they member-declarations).

  • for latter case, i'm not sure why that's necessary. why not directly write down type qualified-id? parameterization necessary?

this about

struct s { typedef int x; }; template <typename t, typename t::x> void f() { } int main() { f<s, 1>(); }

here, typename t means t named template type parameter, typename t::x type of unnamed non-type template parameter.

type-parameter syntax used template type parameters, parameter-declaration syntax used template non-type parameters.

typename t cannot parsed typename-specifier in parameter-declaration, lacks nested-name-specifier, must type-parameter.

typename t::x cannot parsed type-parameter, allows single identifier after typename keyword, must parameter-declaration.

i think there's no ambiguity, text clarifies how differently these 2 template-parameters parsed.


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