c++ - 递归 enable_if 和类型转换

Question

我想将相同数字的指针从 typename U 添加到 typename T，例如当 T = int*** 和U = int*，结果为int****。所以，我写了以下内容:

#include <type_traits>

template <typename T, typename U,
          typename std::enable_if_t<std::is_pointer<U>::value>* = nullptr>
auto addPointer(T, U)
    -> decltype(addPointer(std::declval<std::add_pointer_t<T>>(),
                           std::declval<std::remove_pointer_t<U>>()));

template <typename T, typename U,
          typename std::enable_if_t<!std::is_pointer<U>::value>* = nullptr>
auto addPointer(T, U) -> T;

int main()
{
    using t =
        decltype(addPointer(std::declval<int***>(), std::declval<int*>()));
}

我在 Linux clang 3.7 上得到以下信息:

$ clang++ -std=c++14 -stdlib=libc++ -lc++abi -Wall -Wextra a.cpp 
a.cpp:16:18: error: no matching function for call to 'addPointer'
        decltype(addPointer(std::declval<int***>(), std::declval<int*>()));
                 ^~~~~~~~~~
a.cpp:5:6: note: candidate template ignored: substitution failure [with T = int ***, U =
      int *, $2 = nullptr]: call to function 'addPointer' that is neither visible in the
      template definition nor found by argument-dependent lookup
auto addPointer(T, U)
     ^
/usr/bin/../include/c++/v1/type_traits:244:78: note: candidate template ignored: disabled
      by 'enable_if' [with T = int ***, U = int *]
  ...<bool _Bp, class _Tp = void> using enable_if_t = typename enable_if<_Bp, _Tp>::type;
                                                                         ^
1 error generated.

为什么会出现错误？

Answer 1

当我们处理标量时，ADL 不会在全局命名空间中查找。这样不仅无法找到后备重载，而且无法在 trailing-return-type 中引用您当前定义的重载。

使用 C++14，有一个更好的解决方案可以解决您的问题:

template <typename T>
T addPointer(T, ...);

template <typename T, typename U>
auto addPointer(T t, U* u) {return addPointer(&t, *u);}

Demo .