Matrix operator+ problem

[Dontgiveup]

I have this program that I am using to add 4*4 matrices:

#include<iostream>
using namespace std;
 
class Matrix {           
      
      double x[4][4];
 
      public:
 
      Matrix operator+(Matrix);
 
};
 
Matrix::Matrix operator+(Matrix mat) {
                     
                     Matrix temp;                                
             
                     for(int i = 0; i<4; i++)
                     
                             for(int k = 0; k<4; k++)
                     
                                     temp.x[i][k] = x[i][k] + mat.x[i][k];                     
                             
                     return temp;
}

But the compiler tells me

Matrix::x[4][4] is private within this context

. I don't understand how x can be private to a method that I declared within the class?

[phantomotap]

O_o

You aren't defining a member function.

Soma

[Dontgiveup]

Well isn't the operator+ a member function of the class Matrix? Are operator functions not member methods of a class if declared within that class?

[iMalc]

You aren't defining a member function.

In other words, you've got the Martix:: part in the wrong place, it should be right before 'operator'.

[grumpy]

Change the first line of the definition from

Matrix::Matrix operator+(Matrix mat)

to

Matrix Matrix::operator+(Matrix mat)

That way you will actually define the operator+() you have declared in the class definition, rather than something else.

Generally considered better to pass the argument by (const) reference as well.

[Dontgiveup]

Thanks all. Yeah that was the Problem.

[manasij7479]

btw..

Matrix::Matrix operator+(Matrix mat)

What does that mean to the compiler..?..(as it catches an error relating to a private member....so..this must be somewhat valid...having another meaning)

[whiteflags]

Defines global operator+ I believe. If the compiler tolerates using the scoping operator 'inappropriately' as discussed here. But the implementation did not use public members, obviously.