How to do type convertion for template class?

[Allen]

For example, define vector class as following:

namespace blas {

template<class T> class vector {
....
inline vector<T> operator=()
{
vector<T> t(*this);
return t;
}
};

}

Now define two instances:

blas::vector<double> a;
a = ...;
blas::vector<std::complex<double>> b;
b = a;

When compile the codes, c++ compiler will tell operator = is not
overloaded for type std::complex.
How to solve it?
Thank you very much.

Allen

 

[Rolf Magnus]

For example, define vector class as following:

namespace blas {

template<class T> class vector {
...
inline vector<T> operator=()
{
vector<T> t(*this);
return t;
}

operator= needs an argument - the object to copy from. 'this' is actually
the object to copy _to_.

};

}

Now define two instances:

blas::vector<double> a;
a = ...;
blas::vector<std::complex<double>> b;
b = a;

When compile the codes, c++ compiler will tell operator = is not
overloaded for type std::complex.
How to solve it?

Well, you can make the operator= a template, too.

template<class T> class vector {
...
template<class U>
inline vector<T>& operator=(const vector<U>& rhs)
{
// do your copy operation
return *this;
}
};

 

[Allen]

operator= needs an argument - the object to copy from. 'this' is actually
the object to copy _to_.

Yes, I made a mistake here.

Well, you can make the operator= a template, too.

template<class T> class vector {
...
template<class U>
inline vector<T>& operator=(const vector<U>& rhs)
{
// do your copy operation
return *this;
}

};

It does work now. Thank you.