Question

n3290草案ISO标准：第3.4.2节，第2点

For each argument type T in the function call, there is a set of zero or 
more associated namespaces and aset of zero or more associated classes to
be considered. The sets of namespaces and classes is determined entirely
by the types of the function arguments (and the namespace of any template
template argument).Typedef names and using-declarations used to specify 
the types do not contribute to this set. The sets of namespaces and 
classes are determined in the following way:

— If T is a fundamental type, its associated sets of namespaces and 
  classes are both empty. ##1st point
— If T is a class type (including unions), its associated classes are: 
  the class itself; the class of which it is a member, if any; and its 
  direct and indirect base classes. Its associated namespaces are the 
  namespaces of which its associated classes are members. Furthermore,
  if T is a class template specialization, its associated namespaces and
  classes also include: the namespaces and classes associated with the
  types of the template arguments provided for template type parameters
  (excluding template template parameters); the namespaces of which any
  template template arguments are members; and the classesof which any
  member templates used as template template arguments are members. 
  [ Note: Non-type template arguments do not contribute to the set of
  associated namespaces.—end note ]   ##2nd point
— If T is an enumeration type, its associated namespace is the namespace
  in which it is defined. If it is class member, its associated class is
  the member’s class; else it has no associated class.   ##3rd point
— If T is a pointer to U or an array of U, its associated namespaces and
  classes are those associated with U.    ##4th point
— If T is a function type, its associated namespaces and classes are 
  those associated with the function parameter types and those associated 
  with the return type.        ##5th point 
— If T is a pointer to a member function of a class X, its associated 
  namespaces and classes are those associated with the function parameter 
  types and return type, together with those associated with X.
— If T is a pointer to a data member of class X, its associated namespaces
  and classes are those associated with the member type together with those
  associated with X.       ##6th point

If an associated namespace is an inline namespace (7.3.1), its enclosing 
namespace is also included in the set. If an associated namespace directly
contains inline namespaces, those inline namespaces are also included in
the set. In addition, if the argument is the name or address of a set of
overloaded functions and/or function templates, its associated classes 
and namespaces are the union of those associated with each of the members of
the set, i.e., the classes and namespaces associated with its parameter 
types and return type. Additionally,if the aforementioned set of overloaded
functions is named with a template-id, its associated classes and namespaces
also include those of its type template-arguments and its template
template-arguments.

在这里，我理解了第二，第三和第二第6点......任何人都可以尝试给出一些问题..要理解另一点..请用一个例子解释一下吗？

并且在第二点他给出了一个注释：“非类型模板参数对相关命名空间的集合没有贡献”。解释这也是......？

Answer 1

它解释了如何根据参数查找函数。

1）基本类型不是命名空间或类的成员，因此它们没有贡献。

2b）非类型模板参数是值，不贡献。

4）如果类型是T *或T []，请将其视为只是T。

5）如果参数是另一个函数，则递归搜索其关联的命名空间或类。

来自n3290的一点：依赖于参数的名称查找

1 个答案: