Bringing up an old topic (discussed 15-May), because
it seems noone replied to my post, which contains er..
wrong claims.

Michael Tokarev wrote:
> Shaun Colley wrote:
>>
>> ---
>> if (NULL == (tmp = kmalloc(optlen + 1, GFP_KERNEL))) {
>> retval = -ENOMEM;
>> goto out_unlock;
>> }
>> ---
>>
>> Because kmalloc() takes the 'count' variable as an
>> unsigned number, negative numbers are interpreted as
>> large unsigned numbers. However, if -1 is passed as
>> 'optlen' (represented as 0xffffffff (hex) in unsigned
>> variables, which is the largest value an unsigned
>
> .....
> []
>> And thus, due to the integer overflow, 0 is passed to
>> kmalloc(), causing too little memory to be allocated
>> to hold 'optval'.

> But kmalloc(0) will return NULL, and the whole setsockopt
> will finish with errno set to ENOMEM.
>
> From 2.4 mm/slab.c:
>
> void * kmalloc (size_t size, int flags)
> {
> cache_sizes_t *csizep = cache_sizes;
>
> for (; csizep->cs_size; csizep++) {
> if (size > csizep->cs_size)
> continue;
> return __kmem_cache_alloc(flags & GFP_DMA ?
> csizep->cs_dmacachep : csizep->cs_cachep, flags);
> }
> return NULL;
> }
>
> So, where's the bug?

I was wrong reading the above code, simple as that.
Sure, kmalloc(0) will NOT return NULL as I claimed.

if (size > csizep->cs_size)
continue;
Here, when size == 0 (and csizep->cs_size is always > 0),
the condition is always false, so the next instruction
will be executed, which is:

return __kmem_cache_alloc(flags & GFP_DMA ?
csizep->cs_dmacachep : csizep->cs_cachep, flags);

which will allocate either 32 or 64 bytes of memory (depending
on the arch) and return it to the caller.

So there IS a bug, exactly as described in the original advisory.

I wonder why noone replied... ;)

/mjt

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