pointers...trouble :(

[TreyBrehm]

ok...i am having serveer (<--forgive me for mispelling) difficulty in understanding pointers in the case that its been described...the "&" and the "*" confuse me on a bewildering scale :P if someone could, by any means, try to lighten my burden, it would be greatly appriciated ":D ty

[oz10RXSz]

int x - declaration variable x of type int
int* y - declaration of variable y being a pointer to some variable of type int (pointer to int = int*)
int** z - declaration of variable z being a pointer to a pointer to some variable of type int
and so on

Now, when used in an expression:
x - variable of type int
y - pointer to int
*y - variable of type int pointed by y (you see, in the declaration you had int* y, but it can be also written as int *y, which means that *y must be of type int).
z - pointer to pointer to variable of type int
*z - pointer to variable of type int pointed by z
*(*z) or simply **z - variable of type int pointed by pointer pointed by z :D

*x - invalid, x is not a pointer, so you may not dereference it (* on the left side of a pointer means - take the value pointed by the pointer)
**y or *(*y) - invalid, y is a pointer, but *y is a variable of type int, not a pointer

The & thing is the simplest one. It simply creates a pointer to the thing that is afterwards:
&x - pointer to (the variable x of type int)
&y - pointer to (the pointer y to some int variable)
&z - pointer to (the pointer z to a pointer to int)

So now, when you know how to declare pointers, how to create them and how to use them, you can write:

int x = 5;     // simply a variable of type int
int* y = &x; // 2 things here: declare a new variable y, being a pointer to int and initialize it with a pointer to x
printf("%d\n", *y)  // prints 5 

To summarize: in expressions & is just the opposite than *. & returns a pointer to something.
* returns something pointed by a pointer. And * can be additionally used to declare pointers.

[TreyBrehm]

i THINK that i got it now :D just need a little practice...with the program code that i am currently testing with pointers i am trying to output the PLACE where x and y are stored, as well as the value contained within the x and y variables :P it would let me input the data for some reason...i KNOW this is probably something SO simple but... and help would be appriciated :D Thanks in advance!

[Athar]

int x=5;
int* px=&x;
cout << *px << endl; //prints value of x
cout << px << endl; //prints address of x 

[TreyBrehm]

here is what i have so far...

# include <iostream>
using namespace std;



int main ()
{
int x;
int *ptr1 = &x;
cin>>"enter a value for x ">> x;

int y;
int *ptr2 = &y;
cin>>"enter a value for y ">> y;

cout<<"the location of x is "<< ptr1 << " and the value contained within is "<< *ptr1 << endl;
cout<<"the location of y is "<< ptr2 << " and the value contained within is "<< *ptr2 << endl;
system ("pause");
return 0;
}

[TreyBrehm]

is there no way that i could input the data for the variable that the pointer is pointing to?

[Athar]

The only problem with that code is that you're trying to print something using cin.
Use cout to output text.

[TreyBrehm]

its still not letting me input anything even with the modifications to the code :P
here is what i have changed...

# include <iostream>
using namespace std;
int main ()
{
int x;
int *ptr1 = &x;
cout>>"enter a value for x ";
cin<< x;

int y;
int *ptr2 = &y;
cout>>"enter a value for y ";
cin<< y;

cout<<"the location of x is "<< ptr1 << " and the value contained within is "<< *ptr1 << endl;
cout<<"the location of y is "<< ptr2 << " and the value contained within is "<< *ptr2 << endl;
system ("pause");
return 0;
}

[filipe]

cout << whatever;
cin >> whatever;

[Athar]

You're mixing up input and output operators.
The data should flow in the directions the arrows point to.
Text to cout: cout << text
Value from cin to x: cin >> x

[TreyBrehm]

PERFECT! :D ty