Print Operator overload fail?

Question

Am practicing C++ basic inheritance concepts, came across the need to print out a custom class object and wrote an overload, followed guide exactly yet still does not register use of the new operator for printing (<<).

Am wondering if typed incorrectly or had some declaration/initiation errors somewhere else?

no match for ‘operator<<’ (operand types are ‘std::basic_ostream’ and ‘Choco’) std::cout<< "Choco value: " << child << endl;

#include <iostream>
using namespace std;

class Sweets {
        public:
                // pure virtual, child MUST implement or become abstract
                // Enclosing class becomes automatically abstract - CANNOT instantiate
                virtual bool eat() = 0;
};

// public inheritance : public -> public, protected -> protected, private only accessible thru pub/pro members
class Choco : public Sweets {
        public:

                bool full;

                Choco() {
                        full = false;
                }

                // Must implement ALL inherited pure virtual
                bool eat() {
                        full = true;
                }

                // Overload print operator
                bool operator<<(const Choco& c) {
                        return this->full;
                }
};

int main() {

// Base class object
//sweets parent;
Choco child;

// Base class Ptr
// Ptr value = address of another variable
Sweets* ptr; // pointer to sweets
ptr = &child;

std::cout<< "Sweets* value:  " << ptr << endl;
std::cout<< "Choco address: " << &child << endl;
std::cout<< "Choco value: " << child << endl; // Printing causes error!
}

Related: https://stackoverflow.com/questions/476272/how-to-properly-overload-the-operator-for-an-ostream — Guillaume Racicot, Nov 08 '17 at 20:45
@jamarcus_13 This operator bool operator<<(const Choco& c) { return this->full; } does not make sense at least because the parameter c is not used.:) — Vlad from Moscow, Nov 08 '17 at 20:47
This doesn't address the question, but the logic of this class is inside-out. If a person eats the chocolate, it's the **person** who becomes full, not the **chocolate**. — Pete Becker, Nov 08 '17 at 21:43

Vlad from Moscow · Answer 1 · 2017-11-08T20:54:38.343

0

The operator is defined the following way outside the class because it is not a class member.

std::ostream & operator <<( std::ostream &os, const Choco &c ) 
{
    return os << c.full;
}

Take into account that the function

bool eat() {
    full = true;
}

shall have a return statement with an expression of the type bool or it should be declared with the return type void in the base and derived classes.

edited Nov 08 '17 at 20:54

answered Nov 08 '17 at 20:49

Vlad from Moscow

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Print Operator overload fail?

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