0
include <iostream>
using namespace std;
int main 
{
int m=5;
int* m_pointer;    
m_pointer=&m;
/* &m_pointer != m_pointer ,, thats make sense and understandable
because m_pointer stores the address of (m) but it has it's own address

//_________
int m[5]
// m is const pointer for m[]
// &m[0]= m    make sense
//so why &m = m 
//m stores the address of the array m[] so why it does not have its own address 

}

&m_pointer != m_pointer ,, thats make sense and understandable because m_pointer stores the address of (m) but it has it's own address

Eslam Mohamed
  • 1
  • 1
  • If you look at people queuing up for the latest cool sneakers you will see that the queue starts exactly where the first person in line is standing. – molbdnilo Oct 15 '22 at 15:08
  • exactly its the address of array and the first item together but why m -its like a variable stores the address - have the same address of array and first item u know m=&m=&m[0] and but m must have its own address ,, i assumed that the array name is a constant pointer stores its address like a value ,pls tell me if i was right – Eslam Mohamed Oct 15 '22 at 17:12

0 Answers0