C++ operator= specialization

neutrino17 · 04-23-2008, 03:06 AM

I have a class template parametrized by one typename. I also want to overload its operator= for common case and for double in particular. So I write this code:

Code:

template <typename T>
class field {
    field() {}
    ~field() {}
public:
    T val;
    field<double>& operator =(field<double>& a);
    field<T>& operator =(field<T>& a);
};

template<typename T>
field<T>& field<T>::operator =(field<T>& a)
{
//  common code
    return a;
}
template<>
field<double>& field<double>::operator =(field<double>& a)
{
//  double-specific code
    return a;
}

But g++ says it can not compile that:

Quote:

test.cpp: In instantiation of ‘field<double>’:
test.cpp:23: instantiated from here
test.cpp:17: error: ‘field<T>& field<T>:

perator=(field<T>&) [with T = double]’ cannot be overloaded
test.cpp:12: error: with ‘field<double>& field<T>:

perator=(field<double>&) [with T = double]’

What should I do? I want to be able to run specific code for field<double>:

perator=.

LinuxManMikeC · 04-23-2008, 05:49 AM

Template overloading doesn't work quite the same as function overloading. You don't need to declare the specialized version of the member function in the class definition, only the most generic. Then you just write the specialized implementation as you already have done.

Code:

template <typename T>
class field {
    field() {}
    ~field() {}
public:
    T val;

    //This is wrong, delete!!!
    //field<double>& operator =(field<double>& a);

    field<T>& operator =(field<T>& a);
};

template<typename T>
field<T>& field<T>::operator =(field<T>& a)
{
//  common code
    return a;
}

//This is all you need to create the specialization
template<>
field<double>& field<double>::operator =(field<double>& a)
{
//  double-specific code
    return a;
}

neutrino17 · 04-23-2008, 06:02 AM

Ok, I got it. Thanks.