refrence to a pointer

given a pointer
type * ptr;
When or why would you pass the pointer by refrence to a function ie
function(&ptr);
Instead of simply passing the address directly
function(ptr);

I saw this done and don't understand why.

The prototype for the pass by refrence example
was
function(type **ptr);
A pointer to a pointer. I do not understand.

if the function wanted to change the pointer and not the referenced data it would need to be passed a pointer to the pointer.

Remember that, like any other variable, a pointer itself has an address in memory. When you pass a pointer to a function, a copy of it is made. This copy obviously has a different address than the original. That's why you can do:

Code:

 void print(char * s)
{
    while(*s != 0)
   {
     cout << *(s++);
   }

 cout << endl;
}

The original pointer is unaffected.

Now let's say you wanted to write a function in C that increments a pointer. You would have to do:

Code:

 void inc(char ** p)
{
 *(p++);
}

With C++, it's simply more readable (and less typing) to use:

Code:

 void inc(char * & p)
{
 p++;
}

I got it so given
type *ptr
ptr is the address of the data the pointer points to or in other words is refrenced by
and
&ptr is the address of the pointer which in turn refrences some data.