Pass by reference vs pass by pointer? [duplicate]

Question

Possible Duplicate:
When to pass by reference and when to pass by pointer in C++?

What is the difference between passing by reference and passing the value by a pointer?

Answer 1

When you pass a parameter by reference, the parameter inside the function is an alias to the variable you passed from the outside. When you pass a variable by a pointer, you take the address of the variable and pass the address into the function. The main difference is that you can pass values without an address (like a number) into a function which takes a const reference, while you can't pass address-less values into a function which takes const pointers.

Typically a C++ compiler implement a reference as a hidden pointer.

You can change your function into the pointer variant this way:

void flip(int *i) // change the parameter to a pointer type
{
    cout << "          flip start "<<"i="<< *i<<"\n"; // replace i by *i
    *i = 2*(*i); // I'm not sure it the parenthesis is really needed here,
                 // but IMHO this is better readable
    cout << "          flip exit  "<<"i="<< *i<<"\n";
}

int main()
{
    int j =1;
    cout <<"main j="<<j<<endl;
    flip(&j); // take the address of j and pass this value
    // adjust all other references ...
}

Answer 2

For the second part of your question, here is the code.

#include <iostream>
#include <cassert>

using namespace std;

void flip(int *i)
{
    cout << "          flip start "<<"i="<< i<<"\n";
    *i *= 2;
    cout << "          flip exit  "<<"i="<< i<<"\n";
}

int main()
{
    int j =1;
    cout <<"main j="<<j<<endl;
    flip(&j);
    cout <<"main j="<<j<<endl;
    flip(&j);
    cout <<"main j="<<j<<endl;
    flip(&j);
    cout <<"main j="<<j<<endl;

    assert(j==8);

    return 0;
}

For the first part of your question, I am new to C++ but I find it useful to pass by pointer when having to return multiple outputs for a function. Or to pass NULL as a parameter.