Pointers in C, simple doubts

Question

I'm trying to understand some basic principles of pointers. Someone told me that assigning value to a pointer variable will change the actual variable value. Is that true? I wrote a piece of code and got this:

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int x=5;
    int *address_of_x = &x;
    int y = *address_of_x;
    *address_of_x = 9;
    printf("The value of x is: %d\n", x);
    printf("The X is at: %p\n", &address_of_x);
    printf("value of y = %d\n", y);
    return 0;
}

and got the output like this:

The value of x is: 9
The X is at: 0028FF04
value of y = 5

why the value of "y" stayed 5? Is that because of the ordering of commands?

Answer 1

Yes, it is. address_of_x is assigned a pointer to x , but y is a completely independent int variable. You assign it the same value as x (through a pointer), but x and y are different variables.

At this point, assigning to *address_of_x will change the value of x , but not y .

Answer 2

y isn't a pointer, it is an integer. This line:

int y = *address_of_x;

basically says "take the value pointed to by address_of_x and copy it into y .

If you had instead done this:

int *y = address_of_x;

Then *y would be 9 .

Answer 3

Yes that was because of the ordering of the commands when int y = *address_of_x; this executed the 'address_of_x' contained 5 and hence y got that value

Answer 4

 +--------------+      
 |   5          |
 |*address_of_x |
 +--------------+
        ^
        |         y=*address_of_x =5
        |
 +--------------+
 | address_of_x |
 | 0028FF04     |
 +--------------+

Next time

 *address_of_x = 9

 +--------------+      
 |   9          |
 |*address_of_x |
 +--------------+
        ^
        |         but y still 5
        |
 +--------------+
 | address_of_x |
 | 0028FF04     |
 +--------------+

Answer 5

You are right. Your pointer pointing to x not y . After pointer pointing to x *address_of_x will assigned to y . So y will get the value of 5.

Try to print value of x , It will changed to 9. Because *address_of_x pointing to x .

printf("value of x = %d\n", x); //output = 9

Answer 6

Statement

int y = *address_of_x; 

assigns value at address_of_x to y and then after

  *address_of_x = 9;  

is modifying the variable to which address_of_x points to (which is x here), not y .