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would someone tell me:


s1: int *ptr = (int *) malloc (sizeof(int));

s2: int *ptr = (int *) malloc (n * sizeof(int));

when i use free(ptr),

what is the difference between the two statements?


thanks in advance.

Thomas Zhu wrote:
> would someone tell me:
>
>
> s1: int *ptr = (int *) malloc (sizeof(int));
>
> s2: int *ptr = (int *) malloc (n * sizeof(int));

Please remove the cast. you don't need to cast the value returned by
malloc. Please include stdlib.h.

The difference in the above two statements is the number of bytes
which are marked as reusable.

thanks.

but :

ptr = (int *) malloc (n * sizeof(int));
ptr ++;

free(ptr);

does the system free n mem-units or n-1 mem-units?

and why the cast is not necessary?

Le 21-10-2005, Thomas Zhu <> a écrit*:
> ptr = (int *) malloc (n * sizeof(int));
> ptr ++;
>
> free(ptr);
>
> does the system free n mem-units or n-1 mem-units?

Neither one nor the other. This is UB.

> and why the cast is not necessary?

Because malloc returns a void* pointer, and it
can be implicitely converted into int*.

Marc Boyer

On Fri, 21 Oct 2005 09:37:34 +0000 (UTC),
Marc Boyer <> wrote:


> Le 21-10-2005, Thomas Zhu <> a écrit*:
>> ptr = (int *) malloc (n * sizeof(int));
>> ptr ++;
>>
>> free(ptr);
>>
>> does the system free n mem-units or n-1 mem-units?
>
> Neither one nor the other. This is UB.
>

In this case very likely a painfull UB.

In many implementations the size of an allocated buffer is stored
somwhere just before the buffer itself, and free finds that using a
negative offset from the passed pointer. Obviously, if the ptr given
to free doesn't have the same value as returned from a call to malloc,
free can't find the size of the buffer and thus can't free it properly.

Villy

Thomas Zhu <> wrote:

> ptr = (int *) malloc (n * sizeof(int));
> ptr ++;

> free(ptr);

> does the system free n mem-units or n-1 mem-units?

Neither. If you pass a pointer to free() that was not returned by a
call to malloc(), you get "undefined behavior" - in other words,
absolutely anything may happen at that point.

Furthermore, all you need to know about free() is that it deallocates
all the memory reserved by malloc(); that amount is at least, but by
no means limited to, the amount of memory you asked for.

--
Christopher Benson-Manica | I *should* know what I'm talking about - if I
ataru(at)cyberspace.org | don't, I need to know. Flames welcome.

I''ve got it!!!

I often heard some words (I dont know the their English name , i
translate them from my language to English):
1/memory leak
2/wild pointer

is there any good online books on them ?

thanks a lot.

Thomas Zhu a écrit :
> ptr = (int *) malloc (n * sizeof(int));

What are the words you don't understand in:

"Please remove the cast. you don't need to cast the value returned by
malloc. Please include stdlib.h."

> ptr ++;
>
> free(ptr);

Undefined behaviour.

The value passed to free() must exactly be the value received from malloc().

--
C is a sharp tool

Thanks a lot.

I've got it.
I supposed that the compiler would give a warning to the statement
without a cast.
But I was wrong.
Just now I tried some compilers, they all works.

Thomas Zhu a écrit :
> I often heard some words (I dont know the their English name , i
> translate them from my language to English):
> 1/memory leak

Meaning that some allocated memory can't be freed(). It may happen if
you loose the value of the pointer.

printf ("%p\n", (void *) malloc(123));

or more likely (Ok, strdup() not standard C but is POSIX.1, hence very
portable)

printf ("%s\n", strdup("Hello world"));

> 2/wild pointer

or 'dandling pointer'. An uninitialized pointer or a pointer to an
invalid zone (out of the limits of an array for example). As long as you
don't dereference it, it's fine (well, sort of). But if you dereference
it, it bites (UB).

--
C is a sharp tool