C, pointers, basic problem

Here is my code, just learning more about pointers:
include<stdio.h>
main()
 {
int x, *ptr_x;
int y, *ptr_y;
int z, *ptr_z;
x=64;
y=128;
z=256;
puts("This is a program used to display information on pointers:\n");
printf("x=%d and the pointer is %p\ny=%d and the pointer is %p\nz=%d and the pointer is %p\n",x,y,z,ptr_x,ptr_y,ptr_z);
return 0;
 }

And the output:
This is a program used to display information on pointers:

x=64 and the pointer is 0x80
y=256 and the pointer is 0x8fe07604
z=0 and the pointer is 0x1

Ummm

The pointers were uninitialized.  Sure, they have values, that just happens to be garbage that existed in memory when it began.

Pointers point to locations in memory, right?  I assume you want ptr_x to point to the memory location of variable x, likewise with ptr_y and y, and ptr_z with z.

To do this you do:

ptr_x = &x;
ptr_y = &y;
ptr_z = &z;

the unary & operator before a variable returns the reference (aka memory address) of said variable.

Now ptr_x, ptr_y, and ptr_z should equal the memory locations of variables x, y, and z.

You can "dereference" a pointer by putting a * before it.  So *x_ptr doesn't return x_ptr -- it returns the value of the variable that x_ptr points to (in this case, x).  So you can rewrite the corrected version (with the order that >>4 specified) of your printf statement with just the pointers and their dereference:

printf("x=%d and the pointer is %p\ny=%d and the pointer is %p\nz=%d and the pointer is %p\n", *x_ptr, x_ptr, *y_ptr, y_ptr, *z_ptr, z_ptr);

Likewise, this line:

*x_ptr = 65;

will set the variable x to 65.  But

x_ptr = 65;

will set the memory address (technically it's an offset, but let's not worry about that for now) of x_ptr to 65 and will probably yield unexpected results.  This is one way to introduce memory leaks.