Re: How to know the memory pointed by a ptr is freed?
From: RCollins (rcoll_at_nospam.theriver.com)
Date: 08/19/04
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Date: Thu, 19 Aug 2004 14:10:09 -0700
Keith Thompson wrote:
> RCollins <rcoll@nospam.theriver.com> writes:
> [...]
>
>>No, this is too deep. Given the standard's definition of free(),
>>then I should be able to do this:
>>
>>#include <stdio.h>
>>
>>int mail(void) {
>>unsigned long before, after;
>>int *ptr;
>>
>> ptr = malloc(100); /* ignore possible errors for now */
>> before = (unsigned long) ptr;
>> free(ptr);
>> after = (unsigned long) ptr;
>>
>> if (before == after) printf("bits in ptr have not changed\n")
>> else printf("bits in ptr HAVE changed\n");
>>
>> return 0;
>>}
>
>
> ITYM "main", not "mail" (though of course a function called "mail" is
> perfectly valid). And you're probably assuming that an int* will fit
> into an unsigned long, which isn't guaranteed.
>
> After the call to free(), the value if ptr is indeterminate. Any
> reference to that value, even to convert it to unsigned long, causes
> undefined behavior. On most or all real-world systems, the converson
> just copies the bits, but it could, for example, load the pointer
> value into an address register and trap if the pointer value is
> invalid.
>
> Here's my version of your program. Since refers to the representation
> of ptr, not to its value as a pointer, it doesn't invoke undefined
> behavior.
>
> #include <stdio.h>
> #include <stdlib.h>
> #include <string.h>
> int main(void)
> {
> int *ptr;
> unsigned char *before;
> unsigned char *after;
>
> /*
> * We assume for now that all malloc() calls are successful.
> */
> ptr = malloc(sizeof *ptr);
> before = malloc(sizeof ptr);
> memcpy(before, &ptr, sizeof ptr);
> free(ptr);
> after = malloc(sizeof ptr);
> memcpy(after, &ptr, sizeof ptr);
>
> if (memcmp(before, after, sizeof ptr) == 0) {
> printf("bits in ptr have not changed\n");
> }
> else {
> printf("bits in ptr have changed\n");
> }
> free(before);
> free(after);
>
> return 0;
> }
>
> This program (assuming all the malloc()s succeed) can print "bits in
> ptr have changed" only if there's some compiler magic asociated with
> the free() call that fiddles with ptr, even though the free() function
> has no way within the language to know where ptr is, or even that the
> argument was an object reference.
>
> In my opinion, a conforming compiler cannot perform such magic. It's
> not covered by the "as-if" rule because a program can detect it. A
> *reasonable* program cannot detect it, and I wouldn't complain if the
> standard were amended to allow this particular magic, but I don't
> think that's going to happen.
Agreed ... this was the point of my (flawed) example.
>
>
>>Keep in mind the OP's original question: is there any way to know
>>if the bit representation in "ptr" is a valid memory location?
>
>
> And the answer is: no, there's no portable way to do that. There may
> be non-portable ways to do it, but only by depending on detailed
> knowledge of the implementation (which could easily change with the
> next release of the system).
>
Agreed here, also. I would tend to believe that the desire to
determine if a pointer is "valid" or not is a driving force behind
handle allocation in some OS's (i.e., doubly-indirect memory allocation,
similar to a "pointer to a pointer to an int").
-- Ron Collins Air Defense/RTSC/BCS "I have a plan so cunning, you could put a tail on it and call it a weasel"
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