Why can't I call showA() using A's object?

Question

Why are we able to call the showA() method without object? But if I use void A::showA(A& x) in the method definition then I have to call it using A's object, why?

#include <iostream> 

class A { 

public:
    int a;
    A() { a = 0; } 


     void showA(A&); 
}; 

void showA(A& x) 
{ 

    std::cout << "A::a=" << x.a; 
} 

int main() 
{ 
    A a; 
    showA(a); 
    return 0; 
}

Answer 1

Why are we able to call the showA() method without object?

You don't call the member function A::showA, but instead the free function showA. In fact, the member function A::showA(A&) is declared, but never defined, only the free function showA(A&) has a definition.

If you want to call A::showA, you need a definition;

void A::showA(A& x) { /* ... */ }
//   ^^^ this makes it a member function definition

and then call it as

A a;

a.showA(a);

(Note that it doesn't make much sense to pass the a instance to A::showA invoked on the identical a instance, but that's another issue).

Answer 2

This function

void showA(A& x) 
{ 

    std::cout << "A::a=" << x.a; 
} 

is not a member function of the class A.

It accepts one argument of the type A &.

As for the member function showA then it is declared but not defined.

You could declare it within the class like

class A { 

public:
    int a;
    A() { a = 0; } 


     void showA() const; 
};

and then define it outside the class definition like

void A::showA() const
{ 

    std::cout << "A::a=" << a; 
} 

In this case the function main can look like

int main() 
{ 
    A a; 
    showA(a); 
    a.showA(); 
    return 0; 
}

Answer 3

You can't call it because showA(the one you are thinking) is not the part of the class.It is a global function.The showA function which you declared in class was never defined. In order to do so modify your code a bit. Change this piece of code.

void A::showA(const A& x) { 
std::cout << "A::a=" << x.a; }  // It is advised to make const as it doesn't change state.