pointer doubt....

**roaan** · 07-27-2009

I am a bit puzzled by the following program (found at one of the tutorials on const keyword)

[insert]

Code:

#include <stdio.h>
#include <stdlib.h>

int main(void)
{
	const int my_age = 24;

	printf("\n%d", my_age);
	//my_age = 25; Will generate error

	printf("\n%p", &my_age);
	printf("\n%p", my_age);

	*(int *)(&my_age) = 25;
	printf("\n%d", my_age);
	printf("\n%p", &my_age);
	printf("\n%p", my_age);

	return 0;
}

Though const prefixed variable cannot be changed but it said that using the pointer notation it could be changed. That is fine. My doubt is that i have my_age equal to some integer value say it is 25. And address of my_age is say 0x000012ff.

Now the notation that these guys use is

*(int *)(&my_age) = 25;

I can understand what it is doing. But when i went i thought that even using the following expression should change the value

*(&my_age) = 25; (Value at address of my_age equals to 25, but it does not work)

Why appending (int *) works ? I can understand that its a typecast or may be because by default (&my_age) returns a (void *).

Any thoughts ????????

**System_159** · 07-27-2009

It's because pointers themselves don't have a type, just a location. You must tell the pointer what type the value you're assigning it is(be it an int, a char, a bool, etc...).

**Elysia** · 07-27-2009

No, that's not right. Pointers do have a type.
And the reason it isn't working is because my_age is of type const int, thus if you take its address, we get a pointer to const it (const int*), which if we dereference, we get back to our previous type: const int.
The cast changes the type of the pointer from const int* to int*, then dereferences it. It circumvents the type system, so to speak.

**roaan** · 07-27-2009

And that's why by using the statement

*(&my_age) = 25;

the program doesnt compile even (error : lvalue specifies const object) So i am getting back a const int * which is immutable. But by typecasting to (int *) i convert the const int * to int * and hence it becomes mutable.

I get it now.

@Elysia and above

Thanks ! You people are (awesome * googol times) :-)

**Brafil** · 07-27-2009

Not to mention you shouldn't do this.

**roaan** · 07-27-2009

@Brafil

Not to mention you shouldn't do this.

I didnt get what you meant by that ?

**Elysia** · 07-27-2009

It means you shouldn't "cast away" const, because it's there to protect you from doing mistakes. If you remove the const, then you are no better off than without it.
In fact, you might be off worse than without it because you are promising to not change it, so anyone who reads the code should assume so much. Then what happens when you do change it? Not a good thing.

**roaan** · 07-27-2009

Yes i get it ;-).

**Brafil** · 07-28-2009

That's why they introduced const_cast in c++.

Code:

static_cast<long *>(const_cast<int *>(pointer))

looks cruel (Assuming sizeof(long) == sizeof(int)).

Not like (long *)pointer. Maybe you didn't know that pointer was actually pointing to const ints?

**Elysia** · 07-29-2009

But this is C, not C++.

**Brafil** · 07-29-2009

I know. I just showed an example. You should be careful with any cast, and C++ just makes it easier. But in C, I gotta feeling that you don't have to cast too much.

Just try first without casting. Then, if an error occurs, cast carefully.

**Elysia** · 07-29-2009

Originally Posted by Brafil

But in C, I gotta feeling that you don't have to cast too much.

No, that's the other way around. You cast a lot in C, much less so in C++.

Just try first without casting. Then, if an error occurs, cast carefully.

Now that is an excellent suggestion.
And also double-check twice before you cast that you are sure you know what you're doing and that you are doing the right thing.

**MK27** · 07-29-2009

Originally Posted by Elysia

No, that's the other way around. You cast a lot in C, much less so in C++.

Well, I definitely do

**bithub** · 07-29-2009

No, that's the other way around. You cast a lot in C, much less so in C++.

Not really. C++ is much more picky when it comes to type checking than C is. There's all kinds of dangerous ways to convert between types in C without even needing a cast.

**Elysia** · 07-29-2009

Yes, but good C++ code does not circumvent the type system, so you rarely, if ever, need a cast at all.
And a lot of conversions in C--without casts--generate warnings from the compiler, so casts to shut it up is a good idea.

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