Changing the address of a pointer to edit its value - c++

Question

recently we were trying to use pointer to void, by passing a copy of its address to an instance of a class, in order for the instance to allocate it to the same memory space as an OpenCV Mat it contains, and that is contained in a stack of memory shared by the GPU and CPU to compute it with OpenCV+CUDA.
However while doing that we ran into an issue we didn't quite understand. When passing the address of the pointer to the object, we tried to edit that address so that the pointer was hosted on another space. And while that worked once that copy ran out of scope the value of that pointer returned to being what it was (I'll add a code snip since its confusing to even explain it). However if we accessed the content of the pointer's address that we passed to the function, and edit that, the original pointer is edited.
My hypothesis is that when we passed the pointer's address to the function a copy of the value of that memory address is made, and if that pointer to pointer is edited, the content of it is left undisturbed, while on the other function we accessed the content and directly edited it, hence the copy is left undisturbed and runs out of scope, while the content is edited correctly.
Here is a snip of the code where I try both things:
// Example program
#include <iostream>
#include <string>
void change_value(int **p, int *addr)
{
*p = addr;
}
void change_direction(int **p, int *addr)
{
std::cout << "p value: " << p << std::endl;
std::cout << "p contains: " << *p << std::endl;
std::cout << "p contains contains: " << **p << std::endl;
p = &addr;
std::cout << "p value: " << p << std::endl;
std::cout << "p contains: " << *p << std::endl;
std::cout << "p contains contains: " << **p << std::endl;
}
int main()
{
int a = 1, b = 2, c = 3;
int *ptr = &a;
std::cout << "ptr direction: " << &ptr << std::endl;
std::cout << "ptr value: " << ptr << std::endl;
std::cout << "ptr contains: " << *ptr << std::endl;
std::cout << std::endl;
change_value(&ptr, &b);
std::cout << "change_value" << std::endl;
std::cout << "ptr direction: " << &ptr << std::endl;
std::cout << "ptr value: " << ptr << std::endl;
std::cout << "ptr contains: " << *ptr << std::endl;
std::cout << std::endl;
change_direction(&ptr, &c);
std::cout << std::endl;
std::cout << "change_direction" << std::endl;
std::cout << "ptr direction: " << &ptr << std::endl;
std::cout << "ptr value: " << ptr << std::endl;
std::cout << "ptr contains: " << *ptr << std::endl;
}
This is the output we got. We can see that while ptr is edited on change_value function, since we access to the content of p and edit it, it is not changed on change_direction, since the moment we edit p's address it stops pointing to ptr.
ptr direction: 0x710a27e3fe18
ptr value: 0x710a27e3fe0c
ptr contains: 1
change_value
ptr direction: 0x710a27e3fe18
ptr value: 0x710a27e3fe10
ptr contains: 2
p value: 0x710a27e3fe18
p contains: 0x710a27e3fe10
p contains contains: 2
p value: 0x710a27e3fdd8
p contains: 0x710a27e3fe14
p contains contains: 3
change_direction
ptr direction: 0x710a27e3fe18
ptr value: 0x710a27e3fe10
ptr contains: 2
If anyone could offer a better explanation of what happened I'd be most grateful.
Cheers.

Answer 1

Not sure if I understand the question, but if this:
void change_direction(int **p, int *addr)
{
// ...
p = &addr;
// ...
}
Is supposed to modify the parameter passed to the function, then it isnt correct.
Parameters are passed by value unless you pass them by reference. Passing references is recommended, but if you like you can use pointers. However, then you need to modify what the pointer points to, not the pointer. In the function above p is local to the function. Modifying the value of p has no effect outside the function.
You probably want:
void change_direction(int **p, int *addr)
{
// ...
*p = addr;
// ...
}
Or rather
void change_direction(int *&p, int *addr)
{
// ...
p = addr; // p is a reference, hence no dereference needed
// ...
}
Actually you should prefer int*& here, because a nullptr is not a valid paramter (and thats about the only reason you would use a pointer rather than a reference).

Answer 2

#463035818_is_not_a_number offered a good answer that works well.
But there might be value in fully understanding what your original code is doing.
Beggining from the start:
int a = 1, b = 2, c = 3;
int *ptr = &a;
This initializes three pieces of memory in stack memory containing the int values you gave (a,b,c) and one more piece of memory for your ptr variable that holds the address of a.
You then print out the address and the values of ptr, which obviously doesn't change any of the values.
Next is the call to change_value(int **p, int *addr) which executes.
*p = addr;
The parameter p in this case holds an address to an address to an int somewhere in memory, so *p is the address pointed to by p.
Setting that to addr changes the address p points to.
So in your specific code:
change_value(&ptr,&b);
Sets the value of ptr to be the address of b.
Then next is the change_direction(int **p, int* addr) call (which probably should be named change_address), which executes (ignoring the printing):
p = &addr;
This sets the parameter p to be the address of addr.
But since in C/C++ all functions are called by value (in C there arent even any references, in C++ you have to specify it as #463035818_is_not_a_number mentioned), this doesnt do anything to the original values, that the function has been called with.
So the call:
change_direction(&ptr,&c)
Does not actually change what is held in ptr. At the beginning of the call to the function, the address of ptr (so &ptr) gets copied into a piece of stack-memory, which is then used in the function and freed after the function exits.
I hope this clears things up, pointers can be hard to wrap your head around sometimes.
Cheers.