why address of variable is different to what is returned by pointer?

Question

i am using this manipulate function to get the pointers address but it returns anything else.

#include<iostream>

using namespace std;

int *manipulate( int &b) {
    int *a=&b;
    cout<<&a<<" "<<&b<<" ";     // why memory add of 'a' and 'b' different ?
    return a;         // is it returing pointer which stores address of 'b' ?
}

int main() {
    int x=44;
    cout<<&x<<" "; // gives memory address of x;
    int *ptr=&x;
    cout<<ptr<<" "; //  gives mem address of x;
    int &l=x;
    cout<<&l<<" "; //  gives memory add of x;
    int *c=manipulate(x);  //storing returned pointer
    cout<<&c;  // why address of 'c' different to that of returned pointer?
    return 0;
}

Answer 1

in manipulate function you receive &b by reference.

in this line:

int *a=&b;

you have defined a pointer variable a with value of b address

in this line:

cout<<&a<<" "<<&b<<" ";

you have printed the address of a (the pointer) and address of b

if you want them the same you could:

cout<<a<<" "<<&b<<" ";

and finally you have returned the a (pointer to b)

but in your main

int *c=manipulate(x);

you have created a pointer name c with value of address of x

if you want to print the address of x you could:

cout<<c;