Do I understand pointers correctly?

Ok, I'm trying to see if I understand points correctly. Here is how I understand them:

1) Pointers act as a variable that points to the memory adress of another variable:

Code:

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int a = 10;
int *b= &a;

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2) When dereferenced, you can output the value of the variable the pointer references:

Code:

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std::cout<<*b<<endl;
OUTPUT: 10

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3) When not dereferenced, you can output the memory adress of the variable the point references. Which can later be used to read the memory of a variable from another application (such as in a minesweeper cheat):

Code:

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std::cout<<b<<endl;
OUTPUT: *bunch of hex*

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4) Pointers can be used to pass by reference, so that a variable can be modified by a function without having to return it - allowing for the modification of multiple variable within a function:

Code:

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int inc(int a, int *b)
{
    a++:
    *b = 20;//Is this legal?
    return a;
}
int f = 0;
int g = 1;
inc(f, &g)//f is now 1, g is now 20

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Ok, that's all I can think of right now. However, I want to know if I am right on those four points, and if there is anything else I should know.


EDIT: Yeah yeah, this is the 5 trillionth post on pointers, I know. However, unless I ask myself I won't be sure :)

Yes
Yes
Yes and No - reading another processes memory is far more involved than just declaring a pointer.
Yes.

>> inc(f, &g)//f is now 1, g is now 20
The comment is incorrect. After that call f is still 0 because it is passed by value to the inc function. It would work if you had:

Code:

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f = inc(f, &g)//f is now 1, g is now 20

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It would also work if a was a reference parameter instead.

Quote:

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Originally Posted by Salem

Yes and No - reading another processes memory is far more involved

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Yeah, I understand that it needs more than just a pointer - I was just using that as an example of when one might be used :)

Ok, thanks Salem! Now I can rest easily :)


EDIT:

Quote:

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Originally Posted by Daved

>> inc(f, &g)//f is now 1, g is now 20
The comment is incorrect. After that call f is still 0 because it is passed by value to the inc function. It would work if you had:

Code:

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f = inc(f, &g)//f is now 1, g is now 20

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It would also work if a was a reference parameter instead.

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Ok, I thought I needed to do that, but decided not to for reasons I no longer remember.

Quote:

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1) Pointers act as a variable that points to the memory adress of another variable:

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A variable that is a pointer type stores a value just like any other variable. The value it stores is the address of another variable.

Quote:

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2) When dereferenced, you can output the value of the variable the pointer references:

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When you dereference a pointer variable, you are using the address stored in the pointer variable to access the value at that address in memory.

Quote:

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3) When not dereferenced, you can output the memory adress of the variable the point references.

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Like most variables, you can output the value stored in the pointer, which is an address(although for pointers to type char it is trickier).

Quote:

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1) Pointers act as a variable that points to the memory adress of another variable:
Code:


int a = 10;
int *b= &a;

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Pointers act as a variable that points to a memory adress that may be another variable.

Code:

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WIKI *widget = new Pointer_Wiki_Widget();

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in this case the WIKI type pointer called widget points to an objects adress in memory. Is this really a variable?

Quote:

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Originally Posted by curlious

Pointers act as a variable that points to a memory adress that may be another variable.

Code:

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WIKI *widget = new Pointer_Wiki_Widget();

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in this case the WIKI type pointer called widget points to an objects adress in memory. Is this really a variable?

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That's the point I was forgetting to add.

At any rate, I knew that so as far as I'm concerned, I'm good to go unless someone wants to point out anything else I forgot/didn't know.