How exactly does this work? why does it output 5?

Question

#include <iostream>

using namespace std;

int main()
{
int *p;
int x=5;
p=&x;
cout<<*p;
}

Answer 1

It outputs 5 because you are dereferencing x, which contains the value of 5.

How it works exactly when compiled:

mov DWORD PTR [rbp-12], 5

//we move the value of 5 onto the stack, aka "x = 5"

lea rax, [rbp-12]

//we load the address on the stack where we just stored x, and then store it in the RAX register for operations.

mov QWORD PTR [rbp-8], rax

//now we move the address of "x" onto the stack where variable p will be stored.

And so now, this means p points to the location on the stack where "x" is, and when you deference it, you get the value of x.

Answer 2

here, p is a pointer pointed to null currently.
after that you assigned 5 to x; somewhere in memory x is being saved. so it has a address.

let the address of x is "0X".
and now p is pointing to address of x which is "0X".

since you are printing *p, your program prints the value saved at location "0X" , that's why you got 5.