Call overloaded operator for struct defined in same namespace as class in class method

Question

I have a struct coord defined in the namespace saw. In the same namespace, I have also overloaded the operator<< for coord. In the buildSAW() method of SAWBuilder, I successfully create a coord but when I try to use the overloaded << I get Undefined symbols for architecture x86_64: when I try to compile. The program is able to compile successfully however when I use std::out << coord.toString() << std::endl. How can I successfully access the overloaded operator<< for coord within the method for SAWBuilder?

// saw.hpp    
#ifndef __SAW_HPP__
#define __SAW_HPP__

#include <iostream>
#include <sstream>
#include <string>

using std::string;
using std::ostream;

namespace saw
{

    class SAWBuilder
    {
    public:
        SAWBuilder() {}

        int buildSAW() const;

        static const int SIZE_M = 100;

    };

    struct coord
    {
        int x;
        int y;
        int z;

        string toString();
    };

    ostream& operator<<(ostream& os, const coord& c);

}

#endif

The implementation file:

// saw.cpp
#include "saw.hpp"

#include <iostream>
#include <sstream>
#include <string>

using std::string;
using std::ostringstream;

using std::endl;

namespace saw
{
    int SAWBuilder::buildSAW() const
    {
        int visited[SIZE_M][SIZE_M][SIZE_M] = {}; // initalize to zero

        coord starting_coord = coord{SIZE_M/2, SIZE_M/2, SIZE_M/2};

        std::cout << visited[0][0][0] << std::endl;

        std::cout << starting_coord << std::endl; // <- problem here!

        return 0;
    }

    string coord::toString()
    {
        ostringstream out;

        out << "[" << x << ", " << y << ", " << z << "]";

        return out.str();
    }

    ostream& operator<<(ostream& os, coord& c)
    {
        os << c.toString();
        return os;
    }

}

The main file:

// main.cpp    
#include "saw.hpp"

using saw::SAWBuilder;

int main(int argc, char **argv)
{
    SAWBuilder saw;

    int a = saw.buildSAW();

    return a;
}

The Makefile:

CXX = g++
CXXFLAGS = -Wall -pedantic -std=c++17 -g

Saw: main.o saw.o
    ${CXX} ${CXXFLAGS} -o $@ $^

main.o: saw.hpp
saw.o: saw.hpp

Apart from anything else, names such as `__SAW_HPP__` are reserved in C and C++ - you should not be creating them in your own code. — , Apr 06 '19 at 18:03
[A more detailed discussion of Neil's point](https://stackoverflow.com/questions/228783/what-are-the-rules-about-using-an-underscore-in-a-c-identifier) — user4581301, Apr 06 '19 at 18:45

score 2 · Accepted Answer · answered Apr 06 '19 at 19:02

ostream& operator<<(ostream& os, const coord& c);

is declared, but

ostream& operator<<(ostream& os, coord& c)

is defined instead, making it a different function. Note the missing const. I'd vote to close as a typo if not for

os << c.toString();

which requires coord::toString to be a const function, and probably being the reason the missing const is missing: the const-less version compiled, fooling the asker into thinking it was correct.

So in addition to

ostream& operator<<(ostream& os, const coord& c) // added const
{
    os << c.toString();
    return os;
}

The code also needs

struct coord
{
    int x;
    int y;
    int z;

    string toString() const; // added const
};

and later in the implementation

string coord::toString() const // added const
{
    ostringstream out;

    out << "[" << x << ", " << y << ", " << z << "]";

    return out.str();
}

Call overloaded operator for struct defined in same namespace as class in class method

1 Answers1