Why the address of pointer of an Array is the same of the data stored in that pointer?

Question

If you try that piece of code

#include<stdio.h> int main() { 


// Pointer to an integer 
int *p;  

// Pointer to an array of 5 integers 
int (*ptr)[5];  
int arr[] = { 3, 5, 6, 7, 9 };

// Points to 0th element of the arr. 


// Points to the whole array arr. 
ptr = &arr;  

printf("p = %p, address of P = %p\n", p, &p); 
return 0; }

You will get something like p = 0x7fff8e9b4370, P address = 0x7fff8e9b4340 which means the address of pointer P is something and the data inside it is another

but if you try the same with the pointer of the array like this

#include<stdio.h> int main() { 


// Pointer to an integer 
int *p;  

// Pointer to an array of 5 integers 
int (*ptr)[5];  
int arr[] = { 3, 5, 6, 7, 9 };

// Points to 0th element of the arr. 
p = arr; 

// Points to the whole array arr. 
ptr = &arr;  

printf("arr  = %p, arr address = %p\n", arr, &arr); 
return 0; } 

You will get something like arr = 0x7ffda0a04310, arr address = 0x7ffda0a04310

So how come that a pointer data is the same the pointer Address in memory ?!! when we dereference the address of the arr pointer we should get the number 3 but as i understand from this is the address location 0x7ffda0a04310 in the memory has 0x7ffda0a04310 as a data

so where i am mistaken ?

Answer 1

printing p means print p's value, while p's value is address of a integer number or address of an integer array.

printing &p means you print location of p on memory.

printing arr will show address of first element of the array, it is &arr[0].

printing &arr will show location of arr on memory, it is also adress of first element of the array

So printing p will be different with printing &p. arr and &arr are different types but it will give you the same result.

Answer 2

This is because when you use the symbol of an array, it actually evaluates to &arr (address to first element). So because of this arr and &arr are the same address (but not the same type!).

arr is of type int*

&arr is of type int(*)[5]

The difference is when you do pointer arithmetics. Incrementing arr will go to the address of the next element. So *(arr+1) is essentially the same as arr[1]. Incrementing &arr jumps over the whole array (incrementing a pointer always jumps the whole size of the type)