std::remove_reference not removing the reference

This prints the cout, which means the type of T is int&

    #include <vector>
    #include <sys/types.h>
    using namespace std;
    using var_t = std::variant<int, long, double, std::string>;

    int main(int argc, char* argv[]) {
     std::vector<var_t> vec = {10};
     for(auto v : vec) {
        std::visit([](auto&& arg) {
          using T = decltype(arg);
          // if (std::is_same_v<std::remove_reference<T>, int>) {
          if (std::is_same_v<T, int&>) {
            cout<<"int "<<arg<<endl;
          }
        }, v);
      }
     return 0;
   }

Now if i apply std::remove_reference, it doesn’t print. In fact, it doesn’t matter if I use int, int& or int&&, cout doesn’t get printed

 #include <vector>
 #include <sys/types.h>
 using namespace std;
 using var_t = std::variant<int, long, double, std::string>;
 int main(int argc, char* argv[]) {
    std::vector<var_t> vec = {10};
    for(auto v : vec) {
       std::visit([](auto&& arg) {
         using T = decltype(arg);
         if (std::is_same_v<std::remove_reference<T>, int>) {
           cout<<"int "<<arg<<endl;
         }
       }, v);
     }
  return 0;
 }

Why is this happening?

>Solution :

It’s a typo, but fascinating enough to answer: you forgot _t to extract the type from std::remove_reference.

if (std::is_same_v<std::remove_reference_t<T>, int>) {
//                                      ^~

Why does it compile? Well, std::remove_reference is also a class. It’s perfectly fine thing to pass as std::is_same template argument, so the compiler doesn’t object. But what you really want is the type member of this class (which is also conveniently provided as _t alias).