Inheritance and conversion between 2 classes failed c++ - c++

Question

I have class Signal and Image and both classes have method with same name but different input parameters and output parameter. Is is allowed?
template <class T> class Signal {
public:
Signal<T> zeroPadding(Signal<T>);
}
template <class T> class Image:public Signal<T>
{
public:
Image(void);
~Image(void);
Image(int,int);
Image(int,int,double);
Image(int,int,double,double);
Signal<T> zeroPadding(Image<T>);
template <class T> Image<T>::Image(int width,int height):Signal(width,height) {}
template <class T> Image<T>::Image(int width,int height,double dt):Signal(width,height,dt) {}
template <class T> Image<T>::Image(int width,int height,double dt,double t0 ):Signal(width,height,dt,t0) {}
template <class T> Signal<T> Image<T>::zeroPadding(Image<T> im2){
Image<T> i1 =(*this);
Image<T> i2 =im2;
if (i1.getHeight()==i2.getHeight()){
....
return(im2);
}
}
}
int main() {
Image<int> *a=new Image<int>(2,3);
Image<int> *b=new Image<int>(1,3);
(*a).zeroPadding(*b); //when I try to debug it failed with error error LNK2019:
//unresolved external symbol
//"public: __thiscall Signal<int>::~Signal<int>(void)" (??1?$Signal#H##QAE#XZ)
//referenced in function
//"public: __thiscall Image<int>::~Image<int>(void)" (??1?$Image#H##QAE#XZ)
}
Another question: What about if I want to call in main function (*a).Signal::zeroPadding(*b) how I can convert (*a) and (*b) to be Signal?

Answer 1

Because of how the linker works with templates, your implementation should be on the same file as the declaration. You get LNK2019 because the method is only declared but not defined, the compiler doesn't know what to do with it.
Conversion is possible. You need to explicitly cast a pointer to a or b as Signal.

Answer 2

Yes, it is allowed. If member functions are not declared as virtual, then the functions to be called are determined by the caller's class. In your case, The zeroPadding function is being called by Image type which is the derivative class, so the functions in the Image class will overload the base class.
And your error message
unresolved external symbol
public: __thiscall Signal<int>::~Signal<int>(void)" (??1?$Signal#H##QAE#XZ)referenced in function
public: __thiscall Image<int>::~Image<int>(void)" (??1?$Image#H##QAE#XZ)
implies that you have declared the destructor for Signal and Image template class, but didn't implement them, so there will be a link error;
And for your last question, since *a and *b are the derivative class instances of class Signal, so you can convert them to Signal at ease. Every simple methods of type casting will do that.static_cast<Signal<int> >(*a) or directly use (Signal)(*a) is enough.
Update: Seems like that this is a very old but common met issue about seperate compilation. :)
What is seperate compilation? An easy way to explain, if you put your implementation codes in a different source file with the declaration file in writing a template class, then you are using seperate compilation. There is a keyword export specifically for this scenario, but the compiler support isn't that good.
Suggestion is that to put the implementation codes together into the declaration file or maybe include the implementation file into the header file for template class.
Templates and separate compilation
http://www.cs.uofs.edu/~bi/se516/chapter4/tsld022.htm
http://www.cplusplus.com/forum/beginner/73735/