置换模板朋友功能

Question

print

对于特定类型的上下文（第一个参数）

超负荷也是朋友，

没有向前声明这些上下文

clause.h
1. ？
2. 在这里是班级和朋友函数的定义：

   #include <iostream>
   
   template <typename T> struct clause;
   
   template <typename Context, typename T>
   void print(const Context &, const clause<T> &c) {
     std::cerr << c.i << " friend\n";
   }
   
   // In this simplified code, T is not used, but would normally be.
   template <typename T> struct clause {
   private:
     template <typename Context, typename U>
     friend void print(const Context &, const clause<U> &c);
   
     int i = 8;
   };

我尝试了：

#include "clause.h"

class special_context {};

template <typename T> void print(const special_context &, const clause<T> &c) {
  std::cerr << c.i << " overload\n";
}

int main() {
  clause<int> c;
  print(nullptr, c);
  // print(special_context{}, c); // Fails as the overload is not a friend.
}

我还尝试在班级主体中定义“通用”版本：

// Alternative for clause.h

template <typename T> struct clause { 
  template <typename Context>
  friend void print(const Context &, const clause &c) { 
    std::cerr << c.i << " friend\n";
  } 

  private:
  int i = 8;
};

但现在也失败了，因为现在的过载分辨率对

special_context

。

超负荷也是朋友，

我们做不到。我们必须将超载功能列为朋友。

Answer 1

#include <functional>
#include <iostream>
#include <type_traits>

template <typename T>
struct clause {
 private:
  friend struct print;

  int i = 8;
};

class special_context {};

struct print {
  template <typename Context, typename T>
  void operator()(const Context&, const clause<T>& c) {
    std::cerr << c.i << " friend\n";
  }

  template <typename T>
  void operator()(const special_context&, const clause<T>& c) {
    std::cerr << c.i << " overload\n";
  }
};

int main() {
  clause<int> c;
  std::invoke(print{}, nullptr, c);
  std::invoke(print{}, special_context{}, c);
}

https：//godbolt.org/z/5qznpt3or