I have a template class and want to know, how to get template class variable type when it is used as a template parameter of function. I tried to do the following
#include <iostream>
#include <type_traits>
using namespace std;
template <typename T>
class foo
{
};
template <typename templateClass>
void f()
{
if (is_same<typename templateClass::T, int>::value)
cout << "int";
else if (is_same<typename templateClass::T, double>::value)
cout << "double";
else
cout << "Unknown type";
}
int main()
{
f<foo<double>>();
return 0;
}
This code does not compile, because no type named 'T' in 'foo<double>'
. Then I changed it a little bit:
#include <iostream>
#include <type_traits>
using namespace std;
template <typename T>
class foo
{
public:
using Type = T; //can't write : using T = T;
};
template <typename templateClass>
void f()
{
if (is_same<typename templateClass::Type, int>::value)
cout << "int";
else if (is_same<typename templateClass::Type, double>::value)
cout << "double";
else
cout << "Unknown type";
}
int main()
{
f<foo<double>>();
return 0;
}
Now it works fine, but I had to rename template parameter. Can I get template parameter value without renaming it with using
?
You can create traits to extract that information without changing originel types.
template <typename T>
struct template_parameter;
template <template <typename ...> class C, typename T>
struct template_parameter<C<T>>
{
using type = T;
};
template <typename T>
using template_parameter_t = typename template_parameter<T>::type;
and then
template <typename templateClass>
void f()
{
if constexpr (std::is_same_v<template_parameter_t<templateClass>, int>)
std::cout << "int";
else if constexpr (std::is_same_v<template_parameter_t<templateClass>, double>)
std::cout << "double";
else
std::cout << "Unknown type";
}
The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.