Overriding return type in function template specialization
I would like to specialize a function template such that the return type changes depending on the type of the template argument.
class ReturnTypeSpecialization
{
public:
template<typename T>
T Item();
};
// Normally just return the template type
template<typename T>
T ReturnTypeSpecialization::Item() { ... }
// When a float is specified, return an int
// This doesn't work:
template<float>
int ReturnTypeSpecialization::Item() { ... }
Is this possible? I can't use C++11.
Since the specialization has to agree with the base template on the return type, you can make it so by adding a "return type trait", a struct you can specialize and draw the true return type from:
// in the normal case, just the identity
template<class T>
struct item_return{ typedef T type; };
template<class T>
typename item_return<T>::type item();
template<>
struct item_return<float>{ typedef int type; };
template<>
int item<float>();
Live example.
Note that you might want to stick to the following, so you only need to update the return-type in the item_return
specialization.
template<>
item_return<float>::type foo<float>(){ ... }
// note: No `typename` needed, because `float` is not a dependent type
Do all of the specialization in a worker class and use a simple function as a wrapper that will be specialized implicitly.
#include <iostream>
using std::cout;
// worker class -- return a reference to the given value
template< typename V > struct worker
{
typedef V const & type;
static type get( V const & v ) { return v; }
};
// worker class specialization -- convert 'unsigned char' to 'int'
template<> struct worker<unsigned char>
{
typedef int type;
static type get( unsigned char const & v ) { return v; }
};
// mapper function
template< typename V > typename worker<V>::type mapper( V const & v )
{
return worker<V>::get(v);
}
int main()
{
char a='A';
unsigned char b='B';
cout << "a=" << mapper(a) << ", b=" << mapper(b) << "\n";
}
In this example, the specialization of unsigned char
causes it to be converted to an int
so that cout
will display it as a number instead of as a character, generating the following output...
a=A, b=66
Perhaps you could use the following hack. Given these simple type traits:
template<bool b, typename T, typename U>
struct conditional { typedef T type; };
template<typename T, typename U>
struct conditional<false, T, U> { typedef U type; };
template<typename T, typename U>
struct is_same { static const bool value = false; };
template<typename T>
struct is_same<T, T> { static const bool value = true; };
You could write your class and specialized member function as follows:
class ReturnTypeSpecialization
{
public:
template<typename T>
typename conditional<is_same<T, float>::value, int, T>::type
Item();
};
// Normally just return the template type
template<typename T>
typename conditional<is_same<T, float>::value, int, T>::type
ReturnTypeSpecialization::Item() { return T(); }
// When a float is specified, return an int
template<>
int ReturnTypeSpecialization::Item<float>() { return 1.0f; }
Simple test program (uses C++11 just for verification):
int main()
{
ReturnTypeSpecialization obj;
static_assert(std::is_same<decltype(obj.Item<bool>()), bool>::value, "!");
static_assert(std::is_same<decltype(obj.Item<float>()), int>::value, "!");
}
Here is a live example.
You can do template specializations like so:
template<typename T>
T item() {
return T();
}
template<>
float item<float>() {
return 1.0f;
}