Thread: The heap, using new and delete, and pointers

Ok, so when you use the 'new' keyword, the pointer is kind of like the variable, right? So if you had

int* variable = new int;
*variable = 6;

ok so is that correct (syntax)? what would happen if i tried to increment the address of *variable? like this:

variable++;

could i also do this:

int* variable = new int = 6;

?????

and if i try to assign the pointer an address like this:

variable=x;

instead of:

variable=&x;

then am i saying that the address is whatever x equals, and not pointing to the variable x?

I dont have a compiler on this computer right now so i cant test it out. please answer my questions.

>Ok, so when you use the 'new' keyword, the pointer is kind of l>ike the variable, right?
No, the pointer is still the pointer, it only points to a space you have reserved using new.


>So if you had

>int* variable = new int;
>*variable = 6;

>ok so is that correct (syntax)? what would happen if i tried to >increment the address of *variable? like this:

The pointer would point to the next byte in the memory.. which you most like has not reserved.. (bad idea)
variable++;

>could i also do this:

int* variable = new int = 6;

not sure.. but I dont think so.
>?????

>and if i try to assign the pointer an address like this:

variable=x;

you can't set the value of a pointer to a value of a variable, "level-error" I am assuming that x is not a pointer.

>instead of:

variable=&x;

/Laos

I believe if you difference sp? the pointer like this *variable++ it will work cause when you difference a pointer it gives the value instead of the address but you might wanna check with someone first.

Ryan

Use malloc, free and realloc. <stdlib.h>
They are more comfortable.
in *ptr++ you should know that the *operator has higher
priority than the ++operator.

int* variable = new int = 6;

From compiler point of view, this is valid, provided that you cast 6 from (const int) to (int *), the correct type of variable.

int* variable = new int = (int *) 6;

From human point of view, variable is a pointer to int. It is real variable and has 6 as a value. The assignment is incorrect, as you lose address of reserved memory, which cannot be released then.

variable=x;

am i saying that the address is whatever x equals, and not pointing to the variable x?

Yes, it is true. However, x must be retyped as (int *) x.

Ok, so when you use the 'new' keyword, the pointer is kind of like the variable, right?

Certainly yes. It is variable like any other. You can add, get difference, compare etc. Value of a pointer is an address, not location it points to. Address is actually number and can be safely
casted to (int) on 32 bit machines without any changes in its bit pattern.