using std::conditional to choose member type based on template parameter

Question

Here is my program that attempts to use std::conditional to set the type of a member variable based on the value of an integer template parameter.

#include <stdio.h>
#include <type_traits>
using namespace std;

class cool{
public:
  cool(){ printf("Cool!\n");  }
};

class notcool{
public:
  notcool(){ printf("Not cool!\n");  }
};

template<int N>
class myclass{
public:
  typedef std::conditional<N==5,cool,notcool> thetype;
  thetype theinst;
};

int main(){
  printf("Testing\n");
  myclass<5> myc5;
  myclass<6> myc6;
  printf("Done testing\n");
  return 0; 
} 

I would expect my program to give the following output:

Testing

Cool!

Not cool!

Done testing

Instead, the output is

Testing

Done testing

My compiler is GCC 4.9, and the way I compiled this program was using the command g++ test -std=c++11 -o test

Can anyone tell me why the program does not give the output I expect?

Answer 1

typedef typename std::conditional<N==5,cool,notcool>::type thetype;
//      ^^^^^^^^                                    ^^^^^^

Answer 2

thetype is actually not cool , and also not notcool . But std::conditional .

Yes. std::conditional is a type, that's how most type traits are implemented. If you want to get the underlying type from the conditional, you need to access the underlying type member from std::conditional . That's also how most type traits are implemented: They all have a type member which can be used to access the actual type you want, and not the type trait type.

In C++14, you would use std::conditional_t , which is just a alias for the type member of std::conditional , but because you are using C++11, you need to access it explicitly:

typedef typename std::conditional<N == 5, cool, notcool>::type thetype;