Pointers as function parameters.

[Olysold]

Hi, need some clarification if my understanding of this is right.

#include<iostream>
using std::cin;
using std::cout; using std::endl;

void test (int *x)
{
    *x = 5;
}

int main()
{
    int y;
    cin >> y;
    
    test(&y);

    cout << y << endl;
    
    return 0;
}

The relationship between x and y, is it like *x = &y?

[vlad from moscow]

No, the relation between x and y is

x = &y;

because x has type int *. So it stores only addresses.

On the other hand there is a relation

*x = y;

[Olysold]

Okay I don't get it now, can you please explain what is going on with the code?

[vlad from moscow]

It will be more easy to understand if to write simply without using any function

#include <iostream>

int main()
{
   int y;
   int *x = &y;

   std::cin >> y;

   std::cout << "y = " << y << ", *x = " << *x << std::endl;

   *x = 5; 

   std::cout << "y = " << y << ", *x = " << *x << std::endl;
}

EDIT: I updated some typo.:)

[tcs]

Okay I don't get it now, can you please explain what is going on with the code?

May be I fail to see what you mean?
1. y is assign a value from standard input (cin).
2. Next y will be overwritten with the value 5 (test()).
3. Finally y will be written to standard output (cout).

[Olysold]

Sorry I shouldn't have been so vague and actually explain. This is my thought process so far.

So y is assigned a value, then that value is used as an argument for pointer x.

test(&(y)) //What I think y is getting treated as at the moment.

In the function, pointer x gets dereference and is assigned the value of 5.

Since y is the address of pointer x, and when a pointer is dereferenced it is also accessing the object which it is pointing to, y is also assigned (overwritten) the value of 5 (what I understand about pointers, address and dereference from vlad's code).

Is that what's going on?

[vlad from moscow]

If you are speaking about my code then y is not an address. It is an object that gets its value after statement

std::cin >> y;

x is a pointer that is initialized by the address of object y. So after the declaration

int *x = &y;

x is equal to the address of y that is x == &y.

So y and *x is the same object. That means that after

*x = 5;

y is also equal to 5.

[Olysold]

Alright cool. Just as a final clarification is that what's technically happening with the code I posted?

[vlad from moscow]

Exactly the same as in my code.

[Olysold]

Awesome, that cleared up a lot. Thanks vlad you're really helpful =).