可变参数模板中的模糊运算符[]

Question

我正在尝试编译此示例，其中可变 class 模板继承自可变数量的基，每个基实现不同的 operator[]:

#include <iostream>

template <typename T>
struct Field {
  typename T::value_type storage;

  typename T::value_type &operator[](const T &c) {
    return storage;
  }
};

template<typename... Fields>
struct ctmap : public Field<Fields>... {
};

int main() {
    struct age { typedef int value_type; };
    struct last_name { typedef std::string value_type; };

    ctmap<last_name, age> person;

    person[last_name()] = "Smith";
    person[age()] = 104;
    std::cout << "Hello World!" << std::endl;
    return 0;
}

当我用 gcc (Debian 4.9.2-10) 编译时，出现以下错误

main.cpp: In function ‘int main()’:
main.cpp:22:23: error: request for member ‘operator[]’ is ambiguous
     person[last_name()] = "Smith";
                       ^
main.cpp:7:27: note: candidates are: typename T::value_type& Field<T>::operator[](const T&) [with T = main()::age; typename T::value_type = int]
   typename T::value_type &operator[](const T &c) {
                           ^
main.cpp:7:27: note:                 typename T::value_type& Field<T>::operator[](const T&) [with T = main()::last_name; typename T::value_type = std::basic_string<char>]
main.cpp:23:17: error: request for member ‘operator[]’ is ambiguous
     person[age()] = 104;
                 ^
main.cpp:7:27: note: candidates are: typename T::value_type& Field<T>::operator[](const T&) [with T = main()::age; typename T::value_type = int]
   typename T::value_type &operator[](const T &c) {
                           ^
main.cpp:7:27: note:                 typename T::value_type& Field<T>::operator[](const T&) [with T = main()::last_name; typename T::value_type = std::basic_string<char>]

为什么这么模棱两可？

Answer 1

做你想做的便携方式大致是：

template<class...Ts>
struct operator_index_inherit {};
template<class T0, class T1, class...Ts>
struct operator_index_inherit<T0, T1, Ts...>:
  T0, operator_index_inherit<T1, Ts...>
{
  using T0::operator[];
  using operator_index_inherit<T1, Ts...>::operator[];
};
template<class T0>
struct operator_index_inherit<T0>:
  T0
{
  using T0::operator[];
};

然后：

template<class... Fields>
struct ctmap : operator_index_inherit<Field<Fields>...> {
  using base = operator_index_inherit<Field<Fields>...>;
  using base::operator[];
};

这里我们从每个类型线性继承，using operator[] 在我们的 parents.

如果可以using Field<Fields>::operator[]...;我们就不必这样做了。

必须注意构造函数（我没有注意），但您可能不需要这样做。

live example.

实际出了什么问题取决于我不太确定的标准细节。基本上，您正在以一种复杂的方式混合运算符、继承和重载。即使您的代码符合标准（可能符合也可能不符合），但它的符合性会导致一些编译器死掉。

Answer 2

代码无效，gcc 拒绝它是正确的（虽然 clang 3.6.0 接受它 - 这是一个错误）。查找运算符的规则从 [over.match.oper]:

开始

[...] for a binary operator @ with a left operand of a type whose cv-unqualified version is T1 and a right operand of a type whose cv-unqualified version is T2, three sets of candidate functions, designated member candidates, non-member candidates and built-in candidates, are constructed as follows:
— If T1 is a complete class type or a class currently being defined, the set of member candidates is the result of the qualified lookup of T1::operator@ (13.3.1.1.1); otherwise, the set of member candidates is empty.

成员名称的查找规则是（因为我们正在查找 ctmap<last_name,age>::operator[]），来自 [class.member.lookup]:

The lookup set for f in C, called S(f,C), [...] is calculated as follows:

If C contains a declaration of the name f, [...]

Otherwise (i.e., C does not contain a declaration of f or the resulting declaration set is empty), S(f,C) is initially empty. If C has base classes, calculate the lookup set for f in each direct base class subobject Bi, and merge each such lookup set S(f,B_i) in turn into S(f,C).

The following steps define the result of merging lookup set S(f,B_i) into the intermediate S(f,C):
— [...]
— Otherwise, if the declaration sets of S(f,B_i) and S(f,C) differ, the merge is ambiguous: the new S(f,C) is a lookup set with an invalid declaration set and the union of the subobject sets. In subsequent merges, an invalid declaration set is considered different from any other.
— [...]

基本上 - 我们这里有两个基数 class，都有一个 operator[]。两个声明集不同 - 所以合并是不明确的。消除歧义的方法是引入 using-declarations 将所有基本 class 成员函数引入派生 class，以便初始查找集找到一切。

要缩短您的示例：

struct A { void foo(char) { } };
struct B { void foo(int ) { } };

struct C : A, B { };

struct D : A, B {
    using A::foo;
    using B::foo;
};

有了这个层次结构

C c;
c.foo(4);  // error: ambiguous lookup set for foo()

D d;
d.foo('x') // OK: calls A::foo()

可变参数模板中的模糊运算符[]

ambiguous operator[] in variadic template

c++

operator-overloading

ambiguous

variadic-templates

c++11