when you auto increment the pointer arent you incrementing the value of the "location" of this value? (0x9FFF0)

I'm going to copy something form a good C++ website about incrementing pointers. :

Pointer arithmetic

The C language allows you to perform integer addition or subtraction operations on pointers. If pnPtr points to an integer, pnPtr + 1 is the address of the next integer in memory after pnPtr. pnPtr - 1 is the address of the previous integer before pnPtr.

Note that pnPtr+1 does not return the address after pnPtr, but the next object of the type that pnPtr points to. If pnPtr points to an integer (assuming 4 bytes), pnPtr+3 means 3 integers after pnPtr, which is 12 addresses after pnPtr. If pnPtr points to a char, which is always 1 byte, pnPtr+3 means 3 chars after pnPtr, which is 3 addresses after pnPtr.

When calculating the result of a pointer arithmetic expression, the compiler always multiplies the integer operand by the size of the object being pointed to. This is called scaling.

The following program:

int nValue = 7;
int *pnPtr = &nValue;

cout << pnPtr << endl;
cout << pnPtr+1 << endl;
cout << pnPtr+2 << endl;
cout << pnPtr+3 << endl;

Outputs:

0012FF7C

0012FF80

0012FF84

0012FF88

As you can see, each of these addresses differs by 4 (7C + 4 = 80 in hexadecimal). This is because an integer is 4 bytes on the author’s machine.

The same program using short instead of int:

short nValue = 7;
short *pnPtr = &nValue;

cout << pnPtr << endl;
cout << pnPtr+1 << endl;
cout << pnPtr+2 << endl;
cout << pnPtr+3 << endl;

Outputs:

0012FF7C

0012FF7E

0012FF80

0012FF82

Because a short is 2 bytes, each address differs by 2.

It is rare to see the + and – operator used in such a manner with pointers. However, it is more common to see the ++ or — operator being used to increment or decrement a pointer to point to the next or previous element in an array.