rvalue参数无法在函数重载中解析

Question

如何在main()中编译最后一行？

#include <initializer_list>
#include <type_traits>
#include <functional>

template <typename T>
struct foo {
    foo(std::initializer_list<T>) { }

    template <typename C> struct is_foo : std::false_type { };
    template <typename U> struct is_foo<foo<U>> : std::true_type { };

    template <typename Compare>
    std::enable_if_t<!is_foo<Compare>::value> bar(foo&, Compare comp) {
        bool b = comp(T(), T());  // This line compiles thanks to is_foo<Compare>.
    }

    void bar(foo&& f) { bar(std::forward<foo>(f), std::less<T>()); }

    template <typename... Foos>
    void bar(Foos&&...) { }
};

int main() {
    foo<int> f = {1,2,3};
    f.bar({4,5,6});  // Compiles fine
    f.bar(f,f);  // Compiles fine (thanks to is_foo<Compare>)
    f.bar(f,{4,5,6});  // Won't compile
}

它应该调用foo<T>::bar(Foos&&...)。

Test.cpp:27:17: error: no matching function for call to 'foo<int>::bar(foo<int>&, <brace-enclosed initializer list>)'
  f.bar(f,{4,5,6});  // Won't compile
                 ^
Test.cpp:13:44: note: candidate: template<class Compare> std::enable_if_t<(! foo<T>::is_foo<C>::value)> foo<T>::bar(foo<T>&, Compare) [with Compare = Compare; T = int]
  std::enable_if_t<!is_foo<Compare>::value> bar(foo&, Compare comp) {
                                            ^~~
Test.cpp:13:44: note:   template argument deduction/substitution failed:
Test.cpp:27:17: note:   couldn't deduce template parameter 'Compare'
  f.bar(f,{4,5,6});

Answer 1

不确定你究竟想要什么，但是，如果你打算获得一个可变的bar()函数，它接收零个或多个foo<T>参数（或者可以用来初始化foo<T>的参数），那么...如果你接受了一个限制参数的数量（比如63，在下面的例子中）有一个WF技巧我认为showed one time可以根据您的情况进行调整。

如果您定义了typer模板别名

template <typename T, std::size_t>
using typer = T;

和一个递归的struct proBar

template <typename T, std::size_t N = 64U,
          typename = std::make_index_sequence<N>>
struct proBar;

template <typename T, std::size_t N, std::size_t... Is>
struct proBar<T, N, std::index_sequence<Is...>> : public proBar<T, N-1U>
 {
   using proBar<T, N-1U>::bar;

   void bar (typer<T, Is>... ts)
    { }
 };

template <typename T>
struct proBar<T, 0U, std::index_sequence<>>
 {
    void bar ()
     { }
 };

允许定义你需要的bar()函数，你的模板struct foo成为

template <typename T>
struct foo : public proBar<foo<T>>
 {
   using proBar<foo<T>>::bar;

   foo (std::initializer_list<T>)
    { }

   template <typename Compare>
   auto bar (foo &, Compare comp) -> decltype(comp(T(), T()), void())
    { bool b = comp(T(), T()); }
 };

以下是完整的编译示例

#include <initializer_list>

template <typename T, std::size_t>
using typer = T;

template <typename T, std::size_t N = 64U,
          typename = std::make_index_sequence<N>>
struct proBar;

template <typename T, std::size_t N, std::size_t... Is>
struct proBar<T, N, std::index_sequence<Is...>> : public proBar<T, N-1U>
 {
   using proBar<T, N-1U>::bar;

   void bar (typer<T, Is>... ts)
    { }
 };

template <typename T>
struct proBar<T, 0U, std::index_sequence<>>
 {
    void bar ()
     { }
 };

template <typename T>
struct foo : public proBar<foo<T>>
 {
   using proBar<foo<T>>::bar;

   foo (std::initializer_list<T>)
    { }

   template <typename Compare>
   auto bar (foo &, Compare comp) -> decltype(comp(T(), T()), void())
    { bool b = comp(T(), T()); }
 };



int main()
 {
   foo<int> f = {1, 2, 3};
   f.bar({4, 5, 6});
   f.bar(f, f);
   f.bar(f, {4, 5, 6}); // now compile
 }

Answer 2

这有效。

#include <initializer_list>
#include <type_traits>
#include <functional>

template <typename T>
struct foo {
    foo(std::initializer_list<T>) { }

    template <typename C> struct is_foo : std::false_type { };
    template <typename U> struct is_foo<foo<U>> : std::true_type { };

    template <typename Compare>
    std::enable_if_t<!is_foo<Compare>::value> bar(foo&, Compare comp) {
        bool b = comp(T(), T());  // This line compiles thanks to is_foo<Compare>.
    }

    void bar(foo&& f) { bar(std::forward<foo>(f), std::less<T>()); }

    template <typename... Foos>
    void bar(Foos&&...) { }
};

int main() {
    foo<int> f = { 1,2,3 };
    f.bar({ 4,5,6 });  // Compiles fine
    f.bar(f, f);  // Compiles fine (thanks to is_foo<Compare>)
    f.bar(f, foo<int>({ 4,5,6 }));  // Won't compile
}