I think I'm going to add my own answer in here as well as everyone has done an amazing job but I was really confused at what the point of a pointer to a pointer was. The reason why I came up with this is because I was under the impression that all the values except pointers, were passed by value, and pointers were passed by reference. See the following:
void f(int *x){
printf("x: %d\n", *x);
(*x)++;
}
void main(){
int x = 5;
int *px = &x;
f(px);
printf("x: %d",x);
}
would produce:
x: 5
x: 6
This made my think (for some reason) that pointers were passed by reference as we are passing in the pointer, manipulating it and then breaking out and printing the new value. If you can manipulate a pointer in a function... why have a pointer to a pointer in order to manipulate the pointer to begin with!
This seemed wrong to me, and rightly so because it would be silly to have a pointer to manipulate a pointer when you can already manipulate a pointer in a function. The thing with C though; is everything is passed by value, even pointers. Let me explain further using some pseudo values instead of the addresses.
//this generates a new pointer to point to the address so lets give the
//new pointer the address 0061FF28, which has the value 0061FF1C.
void f(int 0061FF1C){
// this prints out the value stored at 0061FF1C which is 5
printf("x: %d\n", 5);
// this FIRST gets the value stored at 0061FF1C which is 5
// then increments it so thus 6 is now stored at 0061FF1C
(5)++;
}
void main(){
int x = 5;
// this is an assumed address for x
int *px = 0061FF1C;
/*so far px is a pointer with the address lets say 0061FF24 which holds
*the value 0061FF1C, when passing px to f we are passing by value...
*thus 0061FF1C is passed in (NOT THE POINTER BUT THE VALUE IT HOLDS!)
*/
f(px);
/*this prints out the value stored at the address of x (0061FF1C)
*which is now 6
*/
printf("x: %d",6);
}
My main misunderstanding of pointers to pointers is the pass by value vs pass by reference. The original pointer was not passed into the function at all, so we cannot change what address it is pointing at, only the address of the new pointer (which has the illusion of being the old pointer as its pointing to the address the old pointer was pointing to!).
