Thread: i dont understand passing function arugments by refrence ,

Code:

int Function(int &x, int &r, int &p);

int Function(int &x, int &r, int &p)
{
//does this just mean 
int A = x; //what ever is typed in to x in main is stored in A...
                because if it is , then i can do this without int &x, and just do                
                int x, in the parameters and A will still get the x value..
return x + r + p;
}

int main()
{
Function(1,2,3);
return 0;
}

show me a good example then with my code.

Code:

#include <iostream>
#include <string>
using namespace std;

void changeName(string& Name, string newName) // A reference so you can change its value
{
     Name = newName;
}

void printName(string Name) // You don't need to change it so it's not a reference
{
     cout << Name << endl;
}

int main() 
{
    string myName = "William";
    printName(myName);
    changeName(myName, "Bill");
    printName(myName);

    return 0;
}

thanks alot i think i finally understand the concept

Code:

#include <iostream>

using namespace std;

void R(int &R)
{
R = 900;
}

int main ()
{

int total; //total now equals 900

R(total);   
    
return 0;
}

i was always thinking varibles in functions are local and cant be accessed outside of it :o

and i cant change the value of total then the value of R will change will it?

Code:

#include <iostream>

using namespace std;

void R(int &R)
{
R = 900;
}

int main ()
{

int total; //total now equals 900
cout << total << endl;  // does it?

R(total);   
cout << total << endl; // Ah now it does.

return 0;
}

Originally Posted by Anddos

and i cant change the value of total then the value of R will change will it?

R only exist in the scope of the functions, so changing total does nothing to it because it doesn't exist. In fact, at no point do they coexist in the same scope. By passing a reference to an integer, you're now theoretically renaming the variable and placing it into the functions scope. It acts just like the variable in main and anything you do to it will return with it to main.

look at this //it gets the same address as well?

Code:

#include <iostream>
#include <windows.h>

using namespace std;

void R(int &R)
{
R = 900;
}

int main ()
{
cout<<"using refrences"<<endl;
int total = 700;
cout<<"before you change total"<<endl;
cout<<total<< "        " << &total <<endl;
Sleep(5000);
R(total); 
cout<<"after we call the R function"<<endl;
cout<<total<< "      "  << &total <<endl;


return 0;
}

The address never changes in total so I don't understand what you're trying to show there. If you were to output the address of R in the function it would be the same as the address of total. What you have to understand is that memory is always in global scope. So when you pass the address to a function variable, it acts like a reference to the arguement. The same as if I were to pass the memory address into global scope:

Code:

#include <iostream>

int *ptr;   // A global pointer

void func() {
   int foo = *ptr;
   std::cout << foo;
}

int main() {
   int bar = 22;
   ptr = &bar;    // This gives a global reference to the data.
   func();        // In the function, foo is set to the value of that address
                  // and outputted.
   
   return 0;
}

You can think of a reference as an alias or a nickname for another variable. In other words, the variable has two names: the original name and the reference name. So, if you do this:

Code:

int Robert = 10;
int& Bobby = Robert;

both Robert and Bobby refer to the same location in memory, and at that location is the integer 10. So if one name is used to change the value a that location, the other name accesses that same changed value.

Code:

       int: 10
       /     \
      /       \
Robert       Bobby

If you set Bobby equal to 30, what will Robert equal?

If you have a function defined like this:

void myFunc(int& aNum);

and you do call it like this:

myFunc(Robert);

then inside the function aNum will be an alias or nickname for Robert. It's as if you had done this:

int& aNum = Robert;

That means any changes made to aNum will be made to the memory location accessed by Robert.