二进制'operator'：未找到采用'Fraction'类型的右侧操作数的运算符（或没有可接受的转换）

Question

[EDIT] I am writing a template class with type argument T . In one of the functions member of that class, a variable with type T is expected to be written to a std::ofstream object. Everything is fine until I instantiate this class with a custom type argument Fraction . The error occurs although I did overload the operator << .

// collection.h
#include <fstream>
template<typename T>
class Collection
{
public:
    void writeToFile();
private:    
    T val;
};

template<typename T>
inline void Collection<T>::writeToFile()
{
    std::ofstream file("output.txt");
    file << val;

}

// Fraction.cpp
#include <iostream>
std::ostream& operator << (std::ostream& str, const Fraction& f)
{
    std::cout << "Hello";
    return str;
}

Answer 1

New answer:

You answer need to declare operator << with a line like this in Fraction.h , and #include "Fraction.h" before the code that uses it:

std::ostream& operator << (std::ostream& str, const Fraction& f);

The concept of declarations vs definitions is fundamental to C++ (and C), so if you don't understand the distinction, search the web about it now to save yourself any more confusion.

Edit: Old answer:

Are you sure that you are really just doing file << arr[i] and not file << somethingElse << arr[i] ? Because if you do the latter, then the static type of file << somethingElse is likely to be std::ostream& rather than std::ofstream& . In that case, the solution is to change operator<< (..., Fraction) to accept (and return) a general std::ostream& rather than std::ofstream& .

Edit: Another possibility: you need to make sure the declaration of operator<< (..., Fraction) is visible from the place you instantiate Collection<Fraction> (ie the declaraction of operator<< is above it).