C++ no operator "<<" match these operand (inheritance) - c++

Question

I have just recently started class inheritance in c++. While I was making a "Test" program, a error occurred with the cout statement. No clue how to fix it and would be appreciate your response.
#include <iostream>
using namespace std;
class Power{
public:
void isWeak(){
cout << " Weak"<< endl;
}
void isStrong(){
cout << " Strong" << endl;
}
};
class Person:public Power{};
class Person2:public Power{};
int main(){
Person human;
Person2 human2;
cout << "Human is " << human.isWeak() << endl; //error
cout << "Human 2 is " << human2.isStrong() << endl; //error
system("pause");
return 0;
}
the main()'s cout statement has that error between the output and human

Answer 1

Change the functions to
char const *isWeak(){
return " Weak";
}
char const *isStrong(){
return " Strong";
}
As currently defined, both functions have void return type, which means the cout statements within main are trying to print void, which doesn't make sense, and is the cause of the error.

Answer 2

You are attempting to print a void:
cout << "Human is " << human.isWeak() << endl;
is the same as typing
cout << "Human is " << void << endl;
Which will not compile. What you need to do is define your functions in either of the following ways:
class Power
{
public:
std::string isWeak()
{
return std::string(" is weak");
}
std::string isStrong()
{
return std::string(" is strong");
}
};
Or, change your code:
cout << "Human is ";
human.isWeak();
cout << endl;
cout << "Human 2 is ";
human2.isStrong();
cout << endl;

Answer 3

Problem is with isWeak() and isStrong() return type. these two functions return void and you are trying to print it. you can try this-
cout << "Human is " ;
human.isWeak();
cout << endl;
cout << "Human 2 is " ;
human2.isStrong();
cout << endl;

Answer 4

You're trying to 'print' a 'void' statement in cout << "Human is " << human.isWeak() << endl;
You'll need to change your isWeak and isStrong functions to return a std::string/const char* or change the way you call them:
to string:
const char* isWeak() {
return " Weak";
}
// then you can do
cout << "Human is " << human.isWeak() << endl;
Or change the way you call the function:
cout << "Human is ";
human.isWeak();
Your isWeak and isStrong functions are void they do not return anything; calling cout << human.isWeak() is expecting isWeak to return something (an int, string, double, etc.).