Why this code every time gives different output when you run it multiple times?

Question

The shared link and code is from "pointers" tutorials. It displays int score = 5; and that which denotes its memory location. Does that memory location always changes || it repeats || it stay same? Help me understand it better.

Thanks in advance!

Pointers description:
Every variable is a memory location, which has its address defined.
That address can be accessed using the ampersand (&) operator (also called the address-of operator), which denotes an address in memory.

//This outputs the memory address, which stores the variable score.

https://www.sololearn.com/learning/1051/1630/2858/1

#include <iostream>
using namespace std;

int main()
{
int score = 5;
cout << &score << endl;

return 0;
}

Answer 1

Whenever you declare a variable in c++ (or any other programming language) and compile your code, the compiler automatically allocates each variable to a memory location.

These memory locations may change if you rerun the program ( they are stored in RAM, and are deleted when the program closes), but they stay the same throughout the same execution.

So for example:
If you declare int a=3;
And then cout &a multiple times in the same program, it will always print out the same value ( the current variable location in ram).
But once you exit your program and restart, the variable is deleted, and will be declared again in a different place when you rerun your program.

Answer 2

Thanks Mohamad Kamar !

However, my question still remains the same. It is because, I reran the code without exiting.

Answer 3

Thanks for the explanation CarrieForle !
That make sense now. It was completely out of my mind on how code exit automatically.

Answer 4

Shivani 📚 [Less Active] once the code execution is done. The code exit automatically. That is, when the main() gets to the return statement.