C++ ClassRoom -- Lecture No. 3

OK

LESSON: POINTERS

Pointers are variables that hold the (memory) address of another variable within them. You can think of them as reference operators but keep in mind that reference operators & pointers are similar but not the same :P.

For Example:
int a=5;
int *ptr;
ptr=&a;
the last statement
shows us that the pointer
"ptr"
is a reference to "int a".

A pointer is declared
with
*
in front of it

Example:
*myFirstPointerEver

Some Code:

#include <iostream>

using namespace std;

int main() {
    int a=5;
    int *ptr;
    ptr=&a;
    cout << ptr << endl;
}

the first two statements of main()
declares the two variables,
one as "int"
and the other as "integer pointer"
in the third statement
the pointer takes the memory address of the variable
and in the last statement
we print the pointer.

That means that the result will show the memory address of int a
usually it is shown as:
0x.....
lets explain something else,
in the *ptr, the memory address of int a is stored.
So, if we want to print the value of int a.

We can do this:

#include<iostream>

using namespace std;

int main() {
    int a=5;
    int *ptr;
    ptr=&a;
    cout << *ptr << endl;
}

as you can see
in the last statement
there is a
*
in front of the pointer
& hence it means that this code will print the value of int a (i.e. "5").
Now if we want to change the value of int a itself
(without doing this:
a=75;
// or something else)
we can do it through pointer because it has the memory address to int a & as it has the memory address to int a, we can modify the value of int a very easily without much code.

For example:

#include<iostream>

using namespace std;

int main() {
    int a=5;
    int *ptr;
    ptr=&a;
    *ptr=6;
    cout << a << endl;
}

as you can see
in the statement
*ptr=6;
we put this
*
before the name of the pointer
because we want to change the value of int a rather than to change the memory address of int a :P.
And in that code, we tell the compiler to change the value of int a to "6".
Now because this pointer has the memory address of int a, automatically the value of int a will become
6
too.

So, in the last statement it prints:
6

Also please look at this interesting code for a moment:

#include <iostream>

using namespace std;

int main() {
    int a=5;
    int *ptr;
    ptr=&a;
    (*ptr)++;
    cout << a << endl;
    // or
    // cout << *ptr << endl;
}

this will increment the value of int a by 1
but the interesting thing in it are those "(" and ")"
because if we don't put them, it will not work
because "++" has more power than "*" in C++. (What I mean to say is that C++ executes "++" before "*" & hence those brackets much be used to increment the value of that int a)
So, it means that it will increment the value of int a rather than it's memory address.

-- LECTURE ENDED --

-- Note --
I will talk about some strings-pointers in the next lesson.

** The End ;)

** Teacher: Hepic
** Moderator: XRahat2011