推导具有两个以上参数的std::函数

这不会编译function_type_deducer([](){}).describe_me();

如果function_type_deducer不是一个模板，它就会起作用。：）非捕获Lambda（空[]）可以隐式转换为函数指针。遗憾的是，在一些模板论证推导中没有考虑隐含转换。请参阅此问题以了解更多信息（请注意，正如评论所示，我的回答并不完全正确）。

它没有注意到x和y之间有一个小的区别，因为y取一个字符串&，其中x取一个字符串。

这不是函数的问题，这是typeid的问题，正如这个简单的测试代码所示：

template<class T>
void x(void(T)){
    T v;
    (void)v;
}
void f1(int){}
void f2(int&){}
int main(){
    x(f1);
    x(f2);
}

Ideone上的实例。输出：

错误："v"声明为引用，但未初始化

一个简单的修复方法可能是使用标签调度：

#include <type_traits> // is_reference
#include <iostream>
#include <typeinfo>
template<class T>
void print_name(std::true_type){
  std::cout << "reference to " << typeid(T).name();
}
template<class T>
void print_name(std::false_type){
  std::cout << typeid(T).name();
}
template<class T>
void print_name(){
  print_name(typename std::is_reference<T>::type());
}

并调用print_name<NextArg>()而不是typeid(NextArg).name()。

我重新发明轮子了吗？

是的，有点不，你没有。Boost.Function为所有参数提供typedef（argN_type样式），并为其数量提供静态常量arity。然而，您不能简单地通用地访问这些typedef。你需要一种迂回的方式来避免意外地访问不存在的东西。tuple的想法效果最好，但它可以用一种更好的方式编写。这是我曾经写过的东西的修改版本：

#include <tuple>
#include <type_traits>
#include <iostream>
#include <typeinfo>
namespace detail{
template<class T>
std::ostream& print_name(std::ostream& os);
template<class T>
std::ostream& print_pointer(std::ostream& os, std::true_type){
  typedef typename std::remove_pointer<T>:: type np_type;
  os << "pointer to ";
  return print_name<np_type>(os);
}
template<class T>
std::ostream& print_pointer(std::ostream& os, std::false_type){
  return os << typeid(T).name();
}
template<class T>
std::ostream& print_name(std::ostream& os, std::true_type){
  return os << "reference to " << typeid(T).name();
}
template<class T>
std::ostream& print_name(std::ostream& os, std::false_type){
  return print_pointer<T>(os, typename std::is_pointer<T>::type());
}
template<class T>
std::ostream& print_name(std::ostream& os){
  return print_name<T>(os, typename std::is_reference<T>::type());
}
// to workaround partial function specialization
template<unsigned> struct int2type{};
template<class Tuple, unsigned I>
std::ostream& print_types(std::ostream& os, int2type<I>){
  typedef typename std::tuple_element<I,Tuple>::type type;
  print_types<Tuple>(os, int2type<I-1>()); // left-folding
  os << ", ";
  return print_name<type>(os);
}
template<class Tuple>
std::ostream& print_types(std::ostream& os, int2type<0>){
  typedef typename std::tuple_element<0,Tuple>::type type;
  return print_name<type>(os);
}
} // detail::
template<class R, class... Args>
struct function_info{
  typedef R result_type;
  typedef std::tuple<Args...> argument_tuple;
  static unsigned const arity = sizeof...(Args);
  void describe_me(std::ostream& os = std::cout) const{
    using namespace detail;
    os << "I return '"; print_name<result_type>(os);
    os << "' and I take '" << arity << "' arguments. They are: nt'";
    print_types<argument_tuple>(os, int2type<arity-1>()) << "'n";
  }
};

Ideone上的实例。输出：

main:   I return 'i' and I take '2' arguments. They are: 
        'i, pointer to pointer to c'
x:      I return 'Ss' and I take '3' arguments. They are: 
        'i, Ss, c'
y:      I return 'Ss' and I take '3' arguments. They are: 
       'i, reference to Ss, c'

lambda函数的答案链接提供了关键提示：您需要为函数调用运算符获取一个指向成员函数的指针。也就是说，如果T是一个函数对象，则需要查看&T::operator()。使用广义SFINAE，您可以确定此函数调用运算符是否存在。以一种可能有点迂回的方式将这些东西放在一起，会产生以下结果（它与最新版本的gcc和clang一起编译，除了clang还不支持的lambda函数）：

#include <iostream>
#include <sstream>
#include <string>
#include <typeinfo>
#include <functional>
#include <utility>
// -----------------------------------------------------------------------------
struct S {
    void f(int, std::string&, void (*)(int)) {}
    void g(int, std::string&, void (*)(int)) const {}
};
// -----------------------------------------------------------------------------
template <typename T> struct describer;
template <> struct describer<S>;
template <> struct describer<int>;
template <> struct describer<void>;
template <> struct describer<std::string>;
template <typename T> struct describer<T&>;
template <typename T> struct describer<T*>;
template <typename T> struct describer<T const>;
template <typename T> struct describer<T volatile>;
template <typename T> struct describer<T const volatile>;
template <typename T, int Size> struct describer<T(&)[Size]>;
template <typename T> struct describer {
    static std::string type() { return "???"; }
};
template <> struct describer<S> {
    static std::string type() { return "S"; }
};
template <> struct describer<void> {
    static std::string type() { return "void"; }
};
template <> struct describer<int> {
    static std::string type() { return "int"; }
};
template <> struct describer<std::string> {
    static std::string type() { return "std::string"; }
};
template <typename T> struct describer<T&> {
    static std::string type() { return describer<T>::type() + std::string("&"); }
};
template <typename T> struct describer<T&&> {
    static std::string type() { return describer<T>::type() + std::string("&&"); }
};
template <typename T> struct describer<T*> {
    static std::string type() { return describer<T>::type() + std::string("&"); }
};
template <typename T> struct describer<T const> {
    static std::string type() { return describer<T>::type() + std::string(" const"); }
};
template <typename T> struct describer<T volatile> {
    static std::string type() { return describer<T>::type() + std::string(" volatile"); }
};
template <typename T> struct describer<T const volatile> {
    static std::string type() { return describer<T>::type() + std::string(" const volatile"); }
};
template <typename T, int Size> struct describer<T(&)[Size]>
{
    static std::string type() {
        std::ostringstream out;
        out << "(array of " << Size << " " << describer<T>::type() << " objects)&";
        return out.str();
    }
};
template <typename... T> struct description_list;
template <> struct description_list<> { static std::string type() { return std::string(); } }; 
template <typename T> struct description_list<T> { static std::string type() { return describer<T>::type(); } };
template <typename T, typename... S> struct description_list<T, S...> {
    static std::string type() { return describer<T>::type() + ", " + description_list<S...>::type(); }
};
template <typename R, typename... A>
struct describer<R(*)(A...)>
{
    static std::string type() {
        return "pointer function returning " + describer<R>::type() + " and taking arguments"
            + "(" + description_list<A...>::type() + ")";
    }
};
template <typename R, typename S, typename... A>
struct describer<R(S::*)(A...)>
{
    static std::string type() {
        return "pointer to member function of " + describer<S>::type() + " returning " + describer<R>::type() + " "
            "and taking arguments" + "(" + description_list<A...>::type() + ")";
    }
};
template <typename R, typename S, typename... A>
struct describer<R(S::*)(A...) const>
{
    static std::string type() {
        return "pointer to const member function of " + describer<S>::type() + " returning " + describer<R>::type() + " "
            "and taking arguments" + "(" + description_list<A...>::type() + ")";
    }
};
template <typename T> char (&call_op(decltype(&T::operator())*))[1];
template <typename T> char (&call_op(...))[2];
template <typename T> struct has_function_call_operator { enum { value = sizeof(call_op<T>(0)) == 1 }; };
template <typename T>
typename std::enable_if<!has_function_call_operator<T>::value>::type describe(std::string const& what, T)
{
    std::cout << "describe(" << what << ")=" << describer<T>::type() << "n";
}
template <typename T>
typename std::enable_if<has_function_call_operator<T>::value>::type describe(std::string const& what, T)
{
    std::cout << "describe(" << what << ")=function object: " << describer<decltype(&T::operator())>::type() << "n";
}

int f(std::string, std::string const&, std::string&&) { return 0; }
int g(std::string&, std::string const&) { return 0; }
int main()
{
    describe("int", 1);
    describe("f", &f);
    describe("g", &g);
    describe("S::f", &S::f);
    describe("S::g", &S::g);
    describe("mini-lambda", []{}); // doesn't work with clang, yet.
    describe("std::function<int(int(&)[1], int(&)[2], int(&)[4], int(&)[4])>",
             std::function<int(int(&)[1], int(&)[2], int(&)[4], int(&)[4])>());
}