模板参数推导失败，SFINAE

Question

当我编译这段代码时：

#include <type_traits>

template <typename T>
void do_stuff(std::enable_if_t<std::is_integral<T>::value, T> &t) {}

template <typename T>
void do_stuff(std::enable_if_t<std::is_class<T>::value, T> &t) {}

int main() {
    int i = 1;
    do_stuff(i);
    return 0;
}

海湾合作委员会说：

37325975.cpp: In function ‘int main()’:
37325975.cpp:11:15: error: no matching function for call to ‘do_stuff(int&)’
     do_stuff(i);
               ^
37325975.cpp:4:6: note: candidate: template<class T> void do_stuff(std::enable_if_t<std::is_integral<_Tp>::value, T>&)
 void do_stuff(std::enable_if_t<std::is_integral<T>::value, T> &t) {}
      ^
37325975.cpp:4:6: note:   template argument deduction/substitution failed:
37325975.cpp:11:15: note:   couldn't deduce template parameter ‘T’
     do_stuff(i);
               ^
37325975.cpp:7:6: note: candidate: template<class T> void do_stuff(std::enable_if_t<std::is_class<T>::value, T>&)
 void do_stuff(std::enable_if_t<std::is_class<T>::value, T> &t) {}
      ^
37325975.cpp:7:6: note:   template argument deduction/substitution failed:
37325975.cpp:11:15: note:   couldn't deduce template parameter ‘T’
     do_stuff(i);
           ^

我也试过 msvc 2013。

为什么会出现这些错误？

Live Demo

Answer 1

正如编译器所说，该参数类型是不可推导的，因此您需要手动提供模板参数，如下所示：

do_stuff<int>(i);

更好的选择是将 std::enable_if 放在 return 类型或模板参数列表中：

//Return type
template <typename T>
std::enable_if_t<std::is_integral<T>::value>
do_stuff(T &t) {}

template <typename T>
std::enable_if_t<std::is_class<T>::value> 
do_stuff(T &t) {}

//Parameter list
template <typename T, std::enable_if_t<std::is_integral<T>::value>* = nullptr>
void do_stuff(T &t) {}

template <typename T, std::enable_if_t<std::is_class<T>::value>* = nullptr > 
void do_stuff(T &t) {}

这样仍然可以推导出模板参数：

do_stuff(i);

Answer 2

Why do I get these errors?

因为嵌套名称说明符的模板参数推导失败，即 non-deduced contexts。

The nested-name-specifier (everything to the left of the scope resolution operator ::) of a type that was specified using a qualified-id.

// the identity template, often used to exclude specific arguments from deduction
template<typename T> struct identity { typedef T type; };
template<typename T> void bad(std::vector<T> x, T value = 1);
template<typename T> void good(std::vector<T> x, typename identity<T>::type value = 1);
std::vector<std::complex<double>> x;
bad(x, 1.2);  // P1 = std::vector<T>, A1 = std::vector<std::complex<double>>
              // P1/A1: deduced T = std::complex<double>
              // P2 = T, A2 = double
              // P2/A2: deduced T = double
              // error: deduction fails, T is ambiguous
good(x, 1.2); // P1 = std::vector<T>, A1 = std::vector<std::complex<double>>
              // P1/A1: deduced T = std::complex<double>
              // P2 = identity<T>::type, A2 = double
              // P2/A2: uses T deduced by P1/A1 because T is to the left of :: in P2
              // OK: T = std::complex<double>

Answer 3

当编译器尝试解析 do_stuff(int&) 时，它会看到编译器告诉您的两个候选项。但是无法"work backwards"找到满足std::enable_if_t<std::is_integral<T>::value, T> == int的T，也找不到满足std::enable_if_t<std::is_class<T>::value, T> == int.

的T

如TartanLlama's answer中所述，避免这种情况的方法是使参数可推导（例如do_stuff(T&)）并使return类型 或 后续模板参数 依赖于 T.

模板参数推导失败，SFINAE

Template argument deduction failed, SFINAE

c++

templates

sfinae