Value at and Address of Operator in C

Question

#include <iostream>
using namespace std;

int main()
{
    int i = 10;
    int *p;
    p = &i;
    printf("%u\n%u\n%u", *&p, &*p, &i);
}

All will give same answer.

Suppose Adress of i = 100, and since p is pointing to i, it will have value at p = 100, and suppose address of p = 200.

How, to evaluate it?

Answer 1

We have to understand here this way :

a) 'p' is itself a pointer variable . So taking the address of it followed by dereferencing will give back 'p' itself . And 'p' is nothing but  the address of 'i' .

b) Similarly if we dereference 'p' we get 'i' as 'p' contains the address of 'i' . Now when we again put '&' on it means that we are referring to the address of i only and hence in short we are concerned about &i .

c) The same is given as third argument of printf() in direct form.

Hence all of  :  &*p , *&p and &i will give the same value which is the address of 'i' specifically.

Comments

& --  requires an lvalue operand

* -- returns an lvalue

This is also an important point :)

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I solved it as follow:-

*&p

--> *(address of p)

--> *(200) {which means value at address 200 and 200 is an address not a int value}

-->  100 {which of type address not a type of int value}

Now, &*p

--> &(value at (100) )

--> &(10) which means address of 10 here 10 is an integer type value, In memory we have 10 in multiple memory cells, so how it will give the one which is correct??