std::vector 的 std:array 的 SFINAE 无法在 C++11 中编译
SFINAE of a std:array of std::vector doesn't compile in C++11
我用下面的方法写json格式的对象:
#include <array>
#include <vector>
#include <jsoncpp/json/json.h>
//// Write json SFINAE
template <typename T>
struct has_write_json_method {
template <typename U>
static constexpr decltype(std::declval<U>().write_json(), bool()) test(int) { return true; };
template <typename U>
static constexpr bool test(...) { return false; }
static constexpr bool value = test<T>(int());
};
template <class T>
typename std::enable_if<has_write_json_method<T>::value, Json::Value>::type write_json(const T& object) { return object.write_json(); };
template <class T>
typename std::enable_if<!has_write_json_method<T>::value, Json::Value>::type write_json(const T& object);
//// Write json vector
template <class T>
Json::Value write_json(const std::vector<T>& object_v) {
Json::Value output;
for (int i = 0; i < object_v.size(); ++i) { output[i] = write_json<T>(object_v.at(i)); };
return output;
};
//// Write json array
template <class T, std::size_t N>
Json::Value write_json(const std::array<T, N>& object_v) {
Json::Value output;
for (int i = 0; i < object_v.size(); ++i) { output[i] = write_json<T>(object_v.at(i)); };
return output;
};
//// Write json basic
template <class T>
Json::Value write_json_basic(const T& object) {
Json::Value output;
output = object;
return output;
};
template<>
Json::Value write_json<Json::Value>(const Json::Value& object) { return write_json_basic(object); };
template<>
Json::Value write_json<double>(const double& object) { return write_json_basic(object); };
但是,当我尝试写 std::array
的 std::vector
时:
std::array<std::vector<double>, 4> foo_av;
Json::Value output = write_json(foo_av);
它不能在 C++11 中编译:
undefined reference to `std::enable_if<!has_write_json_method<std::vector<double, std::allocator<double> > >::value, Json::Value>::type write_json<std::vector<double, std::allocator<double> > >(std::vector<double, std::allocator<double> > const&)'
更新
使用 std::string 而不是 Json::value 的可重现示例:
#include <array>
#include <vector>
#include <string>
//// Write string SFINAE
template <typename T>
struct has_write_string_method {
template <typename U>
static constexpr decltype(std::declval<U>().write_string(), bool()) test(int) { return true; };
template <typename U>
static constexpr bool test(...) { return false; }
static constexpr bool value = test<T>(int());
};
template <class T>
typename std::enable_if<has_write_string_method<T>::value, std::string>::type write_string(const T& object) { return object.write_string(); };
template <class T>
typename std::enable_if<!has_write_string_method<T>::value, std::string>::type write_string(const T& object);
//// Write string vector
template <class T>
std::string write_string(const std::vector<T>& object_v) {
std::string output;
for (int i = 0; i < object_v.size(); ++i) { output += write_string<T>(object_v.at(i)); };
return output;
};
//// Write string array
template <class T, std::size_t N>
std::string write_string(const std::array<T, N>& object_v) {
std::string output;
for (int i = 0; i < object_v.size(); ++i) { output += write_string<T>(object_v.at(i)); };
return output;
};
//// Write string basic
template <class T>
std::string write_string_basic(const T& object) {
std::string output;
output = object;
return output;
};
template<>
std::string write_string<double>(const double& object) { return write_string_basic(object); };
int main () {
std::array<std::vector<double>, 4> foo_av;
std::string output = write_string(foo_av);
return 0;
}
使用 Godbolt.org 编译时,x86-64 gcc 10.2 失败:
/opt/compiler-explorer/gcc-10.2.0/bin/../lib/gcc/x86_64-linux-gnu/10.2.0/../../../../x86_64-linux-gnu/bin/ld: /tmp/ccAR3MdL.o: in function `std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > write_string<std::vector<double, std::allocator<double> >, 4ul>(std::array<std::vector<double, std::allocator<double> >, 4ul> const&)':
/home/ce/<source>:35: undefined reference to `std::enable_if<!has_write_string_method<std::vector<double, std::allocator<double> > >::value, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >::type write_string<std::vector<double, std::allocator<double> > >(std::vector<double, std::allocator<double> > const&)'
collect2: error: ld returned 1 exit status
这是因为你将T
发送到写字符串方法,但是你传递的参数不匹配T
:
for (int i = 0; i < object_v.size(); ++i) {
// here --------v
output += write_string<T>(object_v.at(i));
}
这里你发送的T
是正确的std::vector<double>
,但是没有write_string<std::vector<double>>()
需要std::vector<double>
。
您希望编译器选择此函数:
template <class T>
std::string write_string(const std::vector<T>& object_v) {
// ...
};
但是不匹配。你看,这里的 T
是向量的模板参数。因此,write_string<std::vector<T>>
收到 std::vector<std::vector<T>>
.
你能做些什么来解决这个问题?解决方案是让编译器推导参数,而不是专门化函数和重载:
for (int i = 0; i < object_v.size(); ++i) {
// no template args --v
output += write_string(object_v.at(i));
}
我用下面的方法写json格式的对象:
#include <array>
#include <vector>
#include <jsoncpp/json/json.h>
//// Write json SFINAE
template <typename T>
struct has_write_json_method {
template <typename U>
static constexpr decltype(std::declval<U>().write_json(), bool()) test(int) { return true; };
template <typename U>
static constexpr bool test(...) { return false; }
static constexpr bool value = test<T>(int());
};
template <class T>
typename std::enable_if<has_write_json_method<T>::value, Json::Value>::type write_json(const T& object) { return object.write_json(); };
template <class T>
typename std::enable_if<!has_write_json_method<T>::value, Json::Value>::type write_json(const T& object);
//// Write json vector
template <class T>
Json::Value write_json(const std::vector<T>& object_v) {
Json::Value output;
for (int i = 0; i < object_v.size(); ++i) { output[i] = write_json<T>(object_v.at(i)); };
return output;
};
//// Write json array
template <class T, std::size_t N>
Json::Value write_json(const std::array<T, N>& object_v) {
Json::Value output;
for (int i = 0; i < object_v.size(); ++i) { output[i] = write_json<T>(object_v.at(i)); };
return output;
};
//// Write json basic
template <class T>
Json::Value write_json_basic(const T& object) {
Json::Value output;
output = object;
return output;
};
template<>
Json::Value write_json<Json::Value>(const Json::Value& object) { return write_json_basic(object); };
template<>
Json::Value write_json<double>(const double& object) { return write_json_basic(object); };
但是,当我尝试写 std::array
的 std::vector
时:
std::array<std::vector<double>, 4> foo_av;
Json::Value output = write_json(foo_av);
它不能在 C++11 中编译:
undefined reference to `std::enable_if<!has_write_json_method<std::vector<double, std::allocator<double> > >::value, Json::Value>::type write_json<std::vector<double, std::allocator<double> > >(std::vector<double, std::allocator<double> > const&)'
更新
使用 std::string 而不是 Json::value 的可重现示例:
#include <array>
#include <vector>
#include <string>
//// Write string SFINAE
template <typename T>
struct has_write_string_method {
template <typename U>
static constexpr decltype(std::declval<U>().write_string(), bool()) test(int) { return true; };
template <typename U>
static constexpr bool test(...) { return false; }
static constexpr bool value = test<T>(int());
};
template <class T>
typename std::enable_if<has_write_string_method<T>::value, std::string>::type write_string(const T& object) { return object.write_string(); };
template <class T>
typename std::enable_if<!has_write_string_method<T>::value, std::string>::type write_string(const T& object);
//// Write string vector
template <class T>
std::string write_string(const std::vector<T>& object_v) {
std::string output;
for (int i = 0; i < object_v.size(); ++i) { output += write_string<T>(object_v.at(i)); };
return output;
};
//// Write string array
template <class T, std::size_t N>
std::string write_string(const std::array<T, N>& object_v) {
std::string output;
for (int i = 0; i < object_v.size(); ++i) { output += write_string<T>(object_v.at(i)); };
return output;
};
//// Write string basic
template <class T>
std::string write_string_basic(const T& object) {
std::string output;
output = object;
return output;
};
template<>
std::string write_string<double>(const double& object) { return write_string_basic(object); };
int main () {
std::array<std::vector<double>, 4> foo_av;
std::string output = write_string(foo_av);
return 0;
}
使用 Godbolt.org 编译时,x86-64 gcc 10.2 失败:
/opt/compiler-explorer/gcc-10.2.0/bin/../lib/gcc/x86_64-linux-gnu/10.2.0/../../../../x86_64-linux-gnu/bin/ld: /tmp/ccAR3MdL.o: in function `std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > write_string<std::vector<double, std::allocator<double> >, 4ul>(std::array<std::vector<double, std::allocator<double> >, 4ul> const&)':
/home/ce/<source>:35: undefined reference to `std::enable_if<!has_write_string_method<std::vector<double, std::allocator<double> > >::value, std::__cxx11::basic_string<char, std::char_traits<char>, std::allocator<char> > >::type write_string<std::vector<double, std::allocator<double> > >(std::vector<double, std::allocator<double> > const&)'
collect2: error: ld returned 1 exit status
这是因为你将T
发送到写字符串方法,但是你传递的参数不匹配T
:
for (int i = 0; i < object_v.size(); ++i) {
// here --------v
output += write_string<T>(object_v.at(i));
}
这里你发送的T
是正确的std::vector<double>
,但是没有write_string<std::vector<double>>()
需要std::vector<double>
。
您希望编译器选择此函数:
template <class T>
std::string write_string(const std::vector<T>& object_v) {
// ...
};
但是不匹配。你看,这里的 T
是向量的模板参数。因此,write_string<std::vector<T>>
收到 std::vector<std::vector<T>>
.
你能做些什么来解决这个问题?解决方案是让编译器推导参数,而不是专门化函数和重载:
for (int i = 0; i < object_v.size(); ++i) {
// no template args --v
output += write_string(object_v.at(i));
}