§10.4/3详细给出了decl-reachable的所有可能的情况，但是我不能完全理解，考虑一下§10.4/6中描述的例子：

Source file "foo.h":
  namespace N {
  struct X {};
  int d();
  int e();
  inline int f(X, int = d()) { return e(); }
  int g(X);
  int h(X);
}

Module M interface:
module;
#include "foo.h"
export module M;
template<typename T> int use_f() {
  N::X x;                       // N​::​X, N, and ​::​ are decl-reachable from use_­f
  return f(x, 123);             // N​::​f is decl-reachable from use_­f,
                                // N​::​e is indirectly decl-reachable from use_­f
                                //   because it is decl-reachable from N​::​f, and
                                // N​::​d is decl-reachable from use_­f
                                //   because it is decl-reachable from N​::​f
                                //   even though it is not used in this call
}
template<typename T> int use_g() {
  N::X x;                       // N​::​X, N, and ​::​ are decl-reachable from use_­g
  return g((T(), x));           // N​::​g is not decl-reachable from use_­g
}
template<typename T> int use_h() {
  N::X x;                       // N​::​X, N, and ​::​ are decl-reachable from use_­h
  return h((T(), x));           // N​::​h is not decl-reachable from use_­h, but
                                // N​::​h is decl-reachable from use_­h<int>
}
int k = use_h<int>();
  // use_­h<int> is decl-reachable from k, so
  // N​::​h is decl-reachable from k

Module M implementation:
module M;
int a = use_f<int>();           // OK
int b = use_g<int>();           // error: no viable function for call to g;
                                // g is not decl-reachable from purview of
                                // module M's interface, so is discarded
int c = use_h<int>();           // OK

为什么N​::​g不能从use_­g声明可达？为什么N​::​h不能从use_­h声明可达，而N​::​h可以从use_­h<int>声明可达？为什么§10.4/（3.2）或§10.4/（3.3）不适用于它们？