I have a templated class Matrix. I want to specialize a function for the type complex, where T can be anything. I have tried this :

```
6 template <typename T>
7 class Matrix {
8 public :
9 static void f();
10 };
11 template<typename T> void Matrix<T>::f() { cout << "generic" << endl; }
12 template<> void Matrix<double>::f() { cout << "double" << endl; }
13 template<typename T> void Matrix<std::complex<T> >::f() { cout << "complex" << endl; }
```

Line 13 does not compile. How can I do that ?

asked Oct 14 '22 06:10
#### Maxime

In lines 11 and 12 you have declaration of explicit specialization for a member of a class template which is allowed by C++ Standard 14.7/3 (14.5.2/2 contains a good example too). In line 13 you are trying to partially specialize a class template and that is not allowed in this form (this is partial specialization because you don't know the whole type `std::complex<T>`

because it is still depends on `T`

). You should partially specialize the whole class.

answered Nov 15 '22 09:11
#### Kirill V. Lyadvinsky

In fact, I found a clever way to do it through Boost. Since I don't want my library to be dependant on Boost, here is the code :

```
template <class T, T val> struct integral_constant
{
typedef integral_constant<T, val> type;
typedef T value_type;
static const T value = val;
};
typedef integral_constant<bool, true> true_type;
typedef integral_constant<bool, false> false_type;
template <typename T> struct is_complex : false_type{};
template <typename T> struct is_complex<std::complex<T> > : true_type{};
template <typename T>
class Matrix {
public :
static void f() { f_( typename is_complex<T>::type() ); }
private :
static void f_( true_type ) { cout << "generic complex" << endl; }
static void f_( false_type ) { cout << "generic real" << endl; }
};
template<> void Matrix<double>::f() { cout << "double" << endl; }
```

This way, I can use function overloading and template to achievement my goal.

answered Nov 15 '22 08:11
#### Maxime

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