Re: [Dev-C++] Variable Declaration

[Daniel K. O. <dan...@ya...>]

Farzan Hajian wrote:

> I have a problem with variable declaration.
> Please tell me what the following declaration means?
>
> int * const * ptr;
>
> I know that "ptr" is a pointer to "int * const". But what "int * 
> const"  means.


I know that Shachar already answered, but I thought that it was a bit 
confusing; so I'll try too.

Considering that T is a type, you can write:
T obj;
T * pobj;

Where "obj" is a variable of type "T", and "pobj" is a pointer to a 
variable of type T.
Now, replace "T" with a type like "int" and you have:
int obj;
int * obj;

But "int *" is also a type; you could write:
typedef int * pint;   /*   now "pint" is a synonym for "int *"   */
pint obj;
pint * pobj;

The typedef "pint" is just a convenience here; if you remove it, you can 
write the equivalent code:
int * obj;
int * * pobj;

So, "obj" is a pointer to "int", and "pobj" is a pointer to pointer to 
"int".
Now if you want to declare the "int" value as constant, you can:
const int * * pobj;
or
int const * * pobj;

Now you can't change the end of the two pointers (the "int" value).
int * a;
pobj = &a;     /* ok, assigning the address of variable "a"  */
*pobj = a;      /* ok, assigning the address stored in "a" */
**pobj = *a;    /*  error, assignment of read-only location   */

You can also declare the variable itself ("pobj") as const, so its value 
can't be changed.
int * a;
int * * const pobj;
pobj = &a;    /*  error, assignment of read-only location */
*pobj = a;     /* ok, assigning the address stored in "a" */
**pobj = *a;  /* ok, assigning the integer pointed by "a"  */

And finally, you can add a const so the value pointed by "pobj" (a 
pointer) can't be changed.
int * a;
int * const * pobj;
pobj = &a;    /* ok, assigning the address of variable "a"  */
*pobj = a;     /*   error, assignment of read-only location   */
**pobj = *a;   /* ok, assigning the integer pointed by "a"  */


---
Daniel K. O.


	
	
		
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