How do I overload << in std::stringstream [duplicate] - c++

Question

I am writing a small matrix library in C++ for matrix operations. However my compiler complains, where before it did not. This code was left on a shelf for 6 months and in between I upgraded my computer from debian etch to lenny (g++ (Debian 4.3.2-1.1) 4.3.2
) however I have the same problem on a Ubuntu system with the same g++.
Here is the relevant part of my matrix class:
namespace Math
{
class Matrix
{
public:
[...]
friend std::ostream& operator<< (std::ostream& stream, const Matrix& matrix);
}
}
And the "implementation":
using namespace Math;
std::ostream& Matrix::operator <<(std::ostream& stream, const Matrix& matrix) {
[...]
}
This is the error given by the compiler:
matrix.cpp:459: error: 'std::ostream&
Math::Matrix::operator<<(std::ostream&,
const Math::Matrix&)' must take
exactly one argument
I'm a bit confused by this error, but then again my C++ has gotten a bit rusty after doing lots of Java those 6 months. :-)

Answer 1

Just telling you about one other possibility: I like using friend definitions for that:
namespace Math
{
class Matrix
{
public:
[...]
friend std::ostream& operator<< (std::ostream& stream, const Matrix& matrix) {
[...]
}
};
}
The function will be automatically targeted into the surrounding namespace Math (even though its definition appears within the scope of that class) but will not be visible unless you call operator<< with a Matrix object which will make argument dependent lookup find that operator definition. That can sometimes help with ambiguous calls, since it's invisible for argument types other than Matrix. When writing its definition, you can also refer directly to names defined in Matrix and to Matrix itself, without qualifying the name with some possibly long prefix and providing template parameters like Math::Matrix<TypeA, N>.

Answer 2

You have declared your function as friend. It's not a member of the class. You should remove Matrix:: from the implementation. friend means that the specified function (which is not a member of the class) can access private member variables. The way you implemented the function is like an instance method for Matrix class which is wrong.

Answer 3

To add to Mehrdad answer ,
namespace Math
{
class Matrix
{
public:
[...]
}
std::ostream& operator<< (std::ostream& stream, const Math::Matrix& matrix);
}
In your implementation
std::ostream& operator<<(std::ostream& stream,
const Math::Matrix& matrix) {
matrix.print(stream); //assuming you define print for matrix
return stream;
}

Answer 4

Assuming that we're talking about overloading operator << for all classes derived from std::ostream to handle the Matrix class (and not overloading << for Matrix class), it makes more sense to declare the overload function outside the Math namespace in the header.
Use a friend function only if the functionality cannot be achieved via the public interfaces.
Matrix.h
namespace Math {
class Matrix {
//...
};
}
std::ostream& operator<<(std::ostream&, const Math::Matrix&);
Note that the operator overload is declared outside the namespace.
Matrix.cpp
using namespace Math;
using namespace std;
ostream& operator<< (ostream& os, const Matrix& obj) {
os << obj.getXYZ() << obj.getABC() << '\n';
return os;
}
On the other hand, if your overload function does need to be made a friend i.e. needs access to private and protected members.
Math.h
namespace Math {
class Matrix {
public:
friend std::ostream& operator<<(std::ostream&, const Matrix&);
};
}
You need to enclose the function definition with a namespace block instead of just using namespace Math;.
Matrix.cpp
using namespace Math;
using namespace std;
namespace Math {
ostream& operator<<(ostream& os, const Matrix& obj) {
os << obj.XYZ << obj.ABC << '\n';
return os;
}
}

Answer 5

In C++14 you can use the following template to print any object which has a T::print(std::ostream&)const; member.
template<class T>
auto operator<<(std::ostream& os, T const & t) -> decltype(t.print(os), os)
{
t.print(os);
return os;
}
In C++20 Concepts can be used.
template<typename T>
concept Printable = requires(std::ostream& os, T const & t) {
{ t.print(os) };
};
template<Printable T>
std::ostream& operator<<(std::ostream& os, const T& t) {
t.print(os);
return os;
}

Answer 6

I would like to simplify this a little with an example that overloads << to print an array.
First pass both the object types around the << operator
create a function to overload the operator as follows.
#include<iostream>
using namespace std;
void operator<<(ostream& os, int arr[]) {
for (int i = 0;i < 10;i++) {
os << arr[i] << " ";
}
os << endl;
}
int main() {
int arr[10] = { 1,2,3,4,5,6,7,8,9,10 };
cout << arr;
}
if cascading of operators is also required make sure to return cout object
in the overloaded function as follows,
#include<iostream>
using namespace std;
ostream& operator<<(ostream& os, int arr[]) {
for (int i = 0;i < 10;i++) {
cout << arr[i] << " ";
}
cout << endl;
return os;
}
int main() {
int arr[10] = { 1,2,3,4,5,6,7,8,9,10 };
int arr2[10] = { 11,22,33,44,55,66,77,88,99,100 };
// cascading of operators
cout << arr << arr2;
}