Address of pointer - c++

Question

I am a newbie in programming. I have a problem that i can't find an understandable answer by myself. I want to find address of pointer by using C++ and C, but two result are different although they have some similar number. Are they still same address ?
adress of g is :0018F9C4
address of g is: 0018F9D3
This is my code:
#include<iostream>
#include<stdio.h>
void main()
{
char g = 'z';
char*p;
p = &g;
std::cout << "adress of g is :" << &p;
printf("\naddress of g is: %p", p);
}

Answer 1

This line display address of p
std::cout << "address of p is :" << &p;
This line display address in p, ie address of g
printf("\naddress of g is: %p", p);
It's normal to have different result.
Try
std::cout << "address of g is :" << static_cast<void*>(p);
printf("\naddress of g is: %p", p);

Answer 2

On the line with std::cout you print out &p...the address of p itself
On the printf line you print out the value of p..no &p
That's different of course.
You should use
std::cout << "adress of g is :" << (void *)p;
printf("\naddress of g is: %p", p);
And why are you mixing cout and printf?

Answer 3

This example may show it in a better way:
#include<iostream>
#include<stdio.h>
int main()
{
char g = 'z';
char*p = &g;
std::cout << "adress of g is :" << static_cast<void*>(p);
std::cout << "\nadress of p is :" << &p;
return 0;
}
Demo
static_cast<void*>(p) is converting a char* to void* for historical purpose. char* is a pointer but it was used as string in C. So the << operator will deal with it as a value not as a pointer. So the trick here is to convert to another pointer type so the << operator will print its actual value which is the address of g.
Now, &p is the address of p not the address of g. In other word it is the address of the place that the address of g is stored in.
(g=='z') is True
(&g==p) is True
(*p==g) is True
(&p==p) is False
(&g=&p) is False

Answer 4

You need to remember that Address of a pointer and Address of what pointer is pointing to are 2 different thing.
&p --> Address of "P"
p --> Address of "g" what pointer is pointing to

Answer 5

Answer: No, they are not the same address, This is clear as the values are different - A program's memory-address pointing to the same location will always be the same.
Explanation:
(tl;dr - &p is the address of p, not the address of g)
char g = 'z' Does exactly what you'd expect - creates a literal constant, and ensures it will be somewhere in memory during your programs execution.
char*p; p = &g; (which can be shortened to char*p = &g) One again, doesn't what you expect, providing the memory-address of the data g into a variable called p ~ Take note however, that p is also a variable, and thus must be held in memory!
The error in your program is during the last two lines of execution:
std::cout << "adress of g is :" << &p;
printf("\naddress of g is: %p", p);
The first line (using std::cout) will present the address of p, As stated previously, p is a pointer, but a pointer is no different from any other variable - it is simply data stored in memory. As such you are at this point creating a temporary pointer to a pointer - AKA char **.
The second line however, does what you expect, outputting the memory-address stored in p.
A correct program would be as follows.
#include<iostream>
#include<stdio.h>
void main()
{
char g = 'z';
char*p = &g;
std::cout << "adress of g is :" << (int)p; //optionally for C++11 use static_cast<int>
printf("\naddress of g is: %p", p);
}
You could also try using the debugger on this example!
#include<iostream>
#include<stdio.h>
#include<cstdint>
void main()
{
char g = 'z';
char*p = &g;
//THIS ASSUMES 32-bit! Note on 64bit systems a pointer is 64-bit, (you can conclude the result for an 8bit or 16bit addressing model)
uint32_t P_value = (uint32_t)p;
std::cout << "adress of g is :" << P_value;
printf("\naddress of g is: %p", p);
}