Method with same name as the arguments results to different output - c++

Question

Why is the method payment1 giving a different outpout then payment2? payment2 has the same argument name then the method name (I hope I'm saying this right?). I mean that the code this->moneysamename+=moneysamename;. Why can't the compiler differentiate the 2 variables?
Here's the full code:
#include <iostream>
using namespace std;
class Person
{
protected:
static int money;
static int moneysamename;
public:
virtual void payment1(float money1) = 0;
virtual void payment2(float moneysamename) = 0;
};
class Worker : public Person
{
public:
virtual void payment1(float moneyanothername)
{
this->money+= moneyanothername;
cout << "Worker: " << money << endl;
};
virtual void payment2(float moneysamename)
{
this->moneysamename+= moneysamename;
cout << "Worker: " << moneysamename << endl;
};
};
void pay1(Person &m, float sum)
{
m.payment1(sum);
}
void pay2(Person &m, float sum)
{
m.payment2(sum);
}
int Person::money = 0;
int Person::moneysamename = 0;
int main()
{
Worker Gaston;
pay1(Gaston, 200);
pay1(Gaston, 300);
pay2(Gaston, 200);
pay2(Gaston, 300);
}
output:
Worker: 200
Worker: 500
Worker: 200
Worker: 300

Answer 1

Your problem is you are not printing the static variable in payment2 but instead you are printing the local function variable
cout << "Worker: " << moneysamename << endl;
Since in the above line you did not use this-> like you do in
this->moneysamename+= moneysamename;
You get the function local moneysamename and not the static member variable. Change the output to
cout << "Worker: " << this->moneysamename << endl;
and you will get the same output.
The reason you do not get this in payment1 is the function paramter has a different name then the static variable. Had you made the function
virtual void payment1(float money)
{
this->money+= money;
cout << "Worker: " << money << endl;
};
You would have had the same results.

Answer 2

virtual void payment2(float moneysamename)
{
this->moneysamename+= moneysamename;
cout << "Worker: " << moneysamename << endl;
};
Here inside payment2, moneysamename will always refer to the function's first argument by C++ scoping rules (the nearest definition is taken). Hence, always the function's argument will be printed instead of the static protected variable of the parent class with the same name.
To print the local parent's variable you have to go to it's scope by cout << "Worker: " << this->moneysamename << endl;