shadow template

[Winsu]

I would like to use T for all my templates...it's supposed to be the same kind so I dont know why it doesnt allow me to use T in my two templates..

#include <iostream>

using namespace std;

template<typename T>
class complex{
	T r;
	T i;
public:
	complex<T>():r(0),i(0){};
	complex<T>(T a):r(a),i(0){};
	complex(T a,T b):r(a),i(b){};
	
	
	complex<T> (const complex<T>& a){		//copy constructor
		r = a.r;
		i = a.i;
	}
	
	complex<T>& operator=(const complex<T> & a){ // copy assignament;
		
		r = a.r;
		i = a.i;
		return *this;
	}
	
	const int get_r(){		//get real part
		int a = r;
		return a;
	}
	
	const int get_i(){		//get imaginarium part
		int a = i;
		return a;
	}
	
	
	template<typename T>
	friend ostream& operator<<(ostream & os,complex<T>& a);
         ////here is the error...if i do the same with N instead of T, It works.
         ///but i dont know the reason
};
	template<typename T>
	ostream& operator<<(ostream& os, complex<T>& a){
		os<<"Parte real "<<a.get_r()<<" "<<"Parte imaginaria "<<a.get_i()<<endl;
		return os;
	}



int main(){
	
	cout<<"TESTEO DE CONSTRUCTOR Y COPY,ASSIGNAMENT, MOVE..ETC"<<endl;
	complex<int> a;
	complex<int> b(1);
	complex<int> c(4,5);
	complex<int> t(a);
	a = c;
	cout<<c<<endl;
	
	complex<char> rojo;
	complex<char> azul(99,87);
	cout<<rojo<<azul<<c;
	
	
	
}
this code launch an error that is 

[ne555]

Because `T' already has a meaning.


http://stackoverflow.com/a/4661372

By the way, ¿why do you need a friend if you are using only the public interface?

[Winsu]

yes, that's true I was gettting get_r and get_i which are public members...I guess I could use the operator . if I declare that function friend..

I dont understande properly what you mean about a meaning...

It's because both typenames are declared in the same scope and then there's a conflict??

[Winsu]

because I can do this..

#include <iostream>

using namespace std;

template<typename T>
class complex{
	T r;
	T i;
public:
	complex<T>():r(0),i(0){};
	complex<T>(T a):r(a),i(0){};
	complex(T a,T b):r(a),i(b){};
	
	
	complex<T> (const complex<T>& a){		//copy constructor
		r = a.r;
		i = a.i;
	}
	
	complex<T>& operator=(const complex<T> & a){ // copy assignament;
		
		r = a.r;
		i = a.i;
		return *this;
	}
	
	const int get_r(){		//get real part
		int a = r;
		return a;
	}
	
	const int get_i(){		//get imaginarium part
		int a = i;
		return a;
	}
	template<typename N>
	friend ostream& operator<<(ostream& os, complex<N>& a);
	
};
	template<typename T>
	ostream& operator<<(ostream& os, complex<T>& a){
		os<<"Parte real "<<a.r<<" "<<"Parte imaginaria "<<a.i<<endl;
		return os;
	}



int main(){
	
	cout<<"TESTEO DE CONSTRUCTOR Y COPY,ASSIGNAMENT, MOVE..ETC"<<endl;
	complex<int> a;
	complex<int> b(1);
	complex<int> c(4,5);
	complex<int> t(a);
	a = c;
	cout<<c<<endl;
	
	complex<char> rojo;
	complex<char> azul(99,87);
	cout<<rojo<<azul<<c;
	
	
	
}

[ne555]

Your example is equivalent to `the extrovert' on the previous link.
In this case the `operator<<' have access to all complex instantiations. By instance, you could operate with `complex<double>' when you are printing a `complex<char>'
It's quite promiscuous.

(note the example in `Taking advantage of the extrovert')


> I dont understande properly what you mean about a meaning...
> It's because both typenames are declared in the same scope and then there's a conflict??
Yes, it's like saying

int foo;
int foo;

[Winsu]

thanks!!