i create ptr pointer and set it to adress of x then print both of the adress why they are different each other ?

#include <stdio.h>
int main()
{
    int *ptr;
    int x = 20;
    ptr = &x;
    printf("adress of x = %p\n", &x);
    printf("adress ptr = %p\n", &ptr);
}

Output :

adress of x = 0x7ffe31da2274
adress of ptr = 0x7ffe31da2278

>Solution :

• &x produces the address of x.
• &ptr produces the address of ptr.
• ptr produces the address stored in ptr, i.e. the address of x.

The addresses of x and ptr are obviously going to be different, since different values are stored in each (the number 20 in x, and the address of x in ptr), as illustrated by the following memory layout:

int x            int* ptr
@ 0x1008         @ 0x1000          <-- Some made up addresses
+-----------+    +-----------+
|        20 |    |    0x1008 |
+-----------+    +-----------+

If you want the value stored in ptr (the address of x), simply use ptr.

printf( "address of x   = %p\n", (void*)&x   );  // 0x1008
printf( "address of ptr = %p\n", (void*)&ptr );  // 0x1000
printf( "address in ptr = %p\n", (void*)ptr  );  // 0x1008