C++ How to call a void function without errors? - c++

Question

This is an assignment of mine that I picked to do but I am not sure how to fix the error message I am getting at cout << contact.getInformation() << endl;without changing the Void function to a different type or changing the main function (which I am trying to avoid). I think my lack of understanding is in how cout and void functions work together. I tried to remove the cout from the function but that did not work and the only way I could make the code run was when I replaced cout << contact.getInformation() << endl; with contact.getInformation() which I am trying to avoid. I just want the inside of the void function to print when I call cout << contact.getInformation() << endl;
Any help is welcome! Thank you!
#include <stdio.h>
#include <iostream>
#include <string>
using namespace std;
class Contact{
public:
Contact(int id, string name, string telephone, int age)
: _id{ id }, _name{ name }, _telephone{ telephone }, _age{ age } {}
int id() { return _id; }
string name() { return _name; }
string telephone() { return _telephone; }
int age() { return _age; }
void getInformation() {
cout << "ID: " + to_string(_id) + "\n" +
"NAME: " + _name + "\n" +
"TEL: " + _telephone + "\n" +
"AGE: " + to_string(_age) + "\n";
}
private:
int _id;
string _name;
string _telephone;
int _age;
};
int main() {
Contact contact{1, "Michael", "555-555-5555", 15};
cout << contact.getInformation() << endl;
}.
EDIT: Thanks all! I see now that it is impossible to do with those restrictions.

Answer 1

The code you've provided have many issues. You can avoid them if you read some good C++ book, my advice is Scott Meyers Effective C++: 55 Specific Ways to Improve Your Programs and Designs.
don't use using directive unless really necessary. In most cases for std namespace - it is not.
Pass function arguments of non primitive type by reference/const reference rather by value or pointer
Understand const keyword and it usage
Understand constructor static initialization bocks
Understand c++ streams
This is how you code should looks like:
#include <iostream>
#include <string>
class Contact {
public:
Contact(int id,const std::string& name,const std::string& telephone, int age):
_id( id ),
_name( name ),
_telephone( telephone ),
_age( age )
{}
int id() const {
return _id;
}
std::string name() const {
return _name;
}
std::string telephone() const {
return _telephone;
}
int age() const {
return _age;
}
private:
int _id;
std::string _name;
std::string _telephone;
int _age;
};
std::ostream& operator<<(std::ostream& to,const Contact& c)
{
to << "ID: " << c.id() << '\n';
to << "NAME: " << c.name() << '\n';
to << "TEL: " << c.telephone() << '\n';
to << "AGE: " << c.age() << '\n';
to.flush();
return to;
}
int main(int argc, const char** argv)
{
Contact contact = {1, "Michael", "555-555-5555", 15};
std::cout << contact << std::endl;
return 0;
}

Answer 2

What you are asking is not possible. The two conditions you have set (i.e. 1. Do not change the void function to another type, and 2. Do not alter the main method) make it impossible to change your code in some other way so as for the main function to produce the intended outcome.
You can either alter your void function to one that returns something 'printable', e.g. a string, or you can keep your void function printing to cout directly, but then change the main function to call this on its own, outside the context of a cout << construct.
(Or, preferably, as has also been pointed in the comments, instead of void, overload the << operator to make cout work with your specific object type)

Answer 3

The name getInformation suggests it should, well, get the information and not print it.
Therefore you probably want this:
string getInformation() {
return "ID: " + to_string(_id) + "\n" +
"NAME: " + _name + "\n" +
"TEL: " + _telephone + "\n" +
"AGE: " + to_string(_age) + "\n";
}
Instead of this:
void getInformation() {
cout << "ID: " + to_string(_id) + "\n" +
"NAME: " + _name + "\n" +
"TEL: " + _telephone + "\n" +
"AGE: " + to_string(_age) + "\n";
}
Not changing main nor getInformation is not possible.