BlueRyse
BlueRyse

Reputation: 29

Class name does not name a type c++

I'm trying to overload the << operator in this class, but the compiler gives me Pila is not a type error (Pila would be stack, the name of the class). GetNElem is another function that I did not include, don't worry.

#include <vector>
#include <iostream>
using namespace std;


template <class T>
class Pila {
 private:
  vector <T> elem;

 public:
  /* Pila(){

  } */

  Pila ( int n ) {
    elem.resize(n);
  }

  void print(ostream & f_out){
    for (int i = 0; i < getNElem(); i++)
      f_out << elem[i] << " ";
    return;
  }


};

ostream& operator << (ostream& f_out, Pila p ){
  p.print(f_out);
  return f_out;
}

Upvotes: 1

Views: 120

Answers (1)

songyuanyao
songyuanyao

Reputation: 172924

Pila is a class template, you need to specify template argument when use it. And you can make operator<< a function template, then

template <typename T>
ostream& operator << (ostream& f_out, Pila<T> p ){
  p.print(f_out);
  return f_out;
}

BTW: It would be better to pass p by reference to avoid copy operation on Pila which constains a std::vector, and make print a const member function.

template <class T>
class Pila {
  ...

  void print(ostream & f_out) const {
    for (int i = 0; i < getNElem; i++)
      f_out << elem[i] << " ";
  }

};

template <typename T>
ostream& operator << (ostream& f_out, const Pila<T>& p ){
  p.print(f_out);
  return f_out;
}

Upvotes: 4

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