Following pointer returned by malloc(0)

Question

I am trying to understand a portion of code. I am leaving out a lot of the code in order to make it simpler to explain, and to avoid unnecessary confusion.

typedef void *UP_T;

void FunctionC(void *pvD, int Offset) {
    unsigned long long int temp;
    void *pvFD = NULL;

    pvFD = pvD + Offset;
    temp = (unsigned long long int)*(int *)pvFD;
}

void FunctionB(UP_T s) {
    FunctionC(s, 8);
}

void FunctionA() {
    char *tempstorage=(char *)malloc(0);
    FunctionB(tempstorage);
}

int main () {
    FunctionA();
    return 0;
}

Like I said, I am leaving out a ton of code, hence the functions that appear useless because they only have two lines of code.

What is temp? That is what is confusing me. When I run something similar to this code, and use printf() statements along the way, I get a random number for pvD, and pvFD is that random number plus eight.

But, I could also be printing the values incorrectly (using %llu instead of %d, or something like that). I am pretty sure it's a pointer to the location in memory of tempstorage plus 8. Is this correct? I just want to be certain before I continue under that assumption.

`char *tempstorage=(char *)malloc(0);` just returns an address which cannot be dereferenced if at all it returns an address. It is implementation defined whether `malloc(0)` should return an address or `NULL`. — Alok Save, Dec 31 '12 at 14:39
[If size is 0, then malloc() returns either NULL, or a unique pointer value that can later be successfully passed to free()](http://www.kernel.org/doc/man-pages/online/pages/man3/malloc.3.html) — Grijesh Chauhan, Dec 31 '12 at 14:44
May be duplicate: http://stackoverflow.com/questions/2022335/whats-the-point-in-malloc0 And a useful link: http://techpreparation.com/c-interview/malloc-valid-memory-address.htm — Grijesh Chauhan, Dec 31 '12 at 14:45
@GrijeshChauhan: This could be an answer. and is a good answer — MOHAMED, Dec 31 '12 at 14:48

score 3 · Accepted Answer · answered Dec 31 '12 at 14:42

3

The standard specifies that malloc(0) returns either NULL or a valid pointer, but that pointer is never to be dereferenced. There aren't any constraints regarding the actual implementation, so you can't rely on the returned pointer being another plus 8.

answered Dec 31 '12 at 14:42

I was mostly curious about the `temp = ...` line though. If pvFD is the location in memory of tempstorage + 8, then its being typecasted to a int, then turned into a pointer, and then being typecasted again into a unsigned long long int? – halexh Dec 31 '12 at 14:52
1

@halexh `temp = (unsigned long long int)*(int *)pvFD;` means: take the `pvFD` address as a pointer to `int`. Dereference it as such, to obtain an `int`, then upcast that `int` to an `unsigned long long`. – Dec 31 '12 at 14:55

Oliver Charlesworth · Answer 2 · 2012-12-31T14:45:28.073

1

It's random in the sense that malloc is typically non-deterministic (i.e. gives different results from run to run).

The result of malloc(0) is implementation-defined (but perfectly valid), you just shouldn't ever dereference it. Nor should you attempt to do arithmetic on it (but this is generally true; you shouldn't use arithmetic to create pointers beyond the bounds of the allocated memory). However, calling free on it is still fine.

edited Dec 31 '12 at 14:45

answered Dec 31 '12 at 14:39

Oliver Charlesworth

267,707
33
569
680

3

*"The result of malloc(0) is well-defined"*, No. It is Implementation defined. – Alok Save Dec 31 '12 at 14:42

Following pointer returned by malloc(0)

2 Answers2

Related