armv5中c的指针分配错误

Question

I have the following code in c:

LinkedList
MemAllocLinkedList_add(MemAllocLinkedList self, void* data)
{
    LinkedList newElement = (LinkedList)
            MemoryAllocator_allocate(self->ma, sizeof(struct sLinkedList));

    if (newElement == NULL)
        return NULL; 

    newElement->data = data;  

    newElement->next = NULL; 

    LinkedList listEnd = LinkedList_getLastElement((LinkedList) self);

    listEnd->next = newElement;

    return newElement;
}

and

char*
MemoryAllocator_allocate(MemoryAllocator* self, int size)
{
    if (((self->currentPtr - self->memoryBlock) + size) <= self->size) {
        char* ptr = self->currentPtr;
        self->currentPtr += size;
        return ptr;
    }
    else{
        printf("MemoryAllocator_allocate: Out of Memory\n");
        return NULL;
    }
}

being LinkedList a pointer to sLinkedList that is:

struct sLinkedList {
    void* data;
    struct sLinkedList* next;
};

and MemAllocLinkedList a pointer to sMemAllocLinkedList that is:

struct sMemAllocLinkedList {
    void* data;
    struct sLinkedList* next;
    MemoryAllocator* ma;
};

I have a program who calls this 'MemAllocLinkedList_add' function several times without a problem, but there's one point where, within this function, the assignment 'listEnd->next = newElement' switches the first 16 bits with the las 16, so instead of having on listEnd->next 975912 I have -467140594 which leads to a SISEGEV error when I try to access to last element of that list later.

If I compile it and run it on a virtual machine with debian (intel 64bits) and on a raspberry pi (armv6 32 bits) it works perfectly fine. But when I try on a NanosG20 with armv5tej it does this thing that I've explained. I use gcc 4.6.

Anybody knows why this happens?

Thanks.

EDIT:

Here´s LinkedList_getLastElement :

LinkedList
LinkedList_getLastElement(LinkedList list)
{int i=0;
    while (list->next != NULL) {
        list = list->next; i++;     
    }
    return list;
}

Type MemoryAllocator:

typedef struct {
    char* memoryBlock;
    char* currentPtr;
    int size;
} MemoryAllocator;

This code is from libiec61850 an implementation of the standard 61850 for subelectric stations so there is a lot more code which is impossible to post here. I'm just posting the part where I get the error, but like I said, only on a board with ARMv5, not on the VM or the Raspberry Pi.

Answer 1

You actually have the same problem on all 3 platforms, it's just that on Intel and ARMv6 you get the "appearing to work correctly" form of undefined behaviour.

Here's a relevant quote from the C standard (6.3.2.3 in the n1256 draft of C99 I have to hand):

A pointer to an object or incomplete type may be converted to a pointer to a different object or incomplete type. If the resulting pointer is not correctly aligned for the pointed-to type, the behavior is undefined.

The notion of "correctly aligned" is defined by the ABI for the platform. In this case, the ARM ABI says thar char requires 1-byte alignment, and pointers require 4-byte alignment. Since struct sLinkedList consists of two pointers, it must also be aligned at least as strictly as a pointer.

Now, since MemoryAllocator_allocate naïvely increments its internal free pointer by size each time, there's no guarantee that the char * it returns is anything better than 1-byte aligned, thus converting that to a struct sLinkedList * leads to undefined behaviour in 3/4 possible cases. Since compilers tend to be more pragmatic than vindictive, rather than nasal demons you got the undefined behaviour of "whatever the hardware does for unaligned loads/stores" upon dereferencing it, which may be an alignment fault, returning some nonsense data, or even returning the data you'd expect.

The solution is to make MemoryAllocator_allocate always round up size to the nearest multiple of the largest required alignment - for the ARM ABI, this is 8 bytes, although if you're sure you will never use any 64-bit types (like double or long long ), you can probably get away with 4 bytes.