[PATCH v2 01/50] lib: Add memdup()
Sean Anderson
sean.anderson at seco.com
Thu May 6 19:57:31 CEST 2021
On 5/6/21 1:41 PM, Simon Glass wrote:
> Hi Pratyush,
>
> On Thu, 6 May 2021 at 10:07, Pratyush Yadav <p.yadav at ti.com> wrote:
>>
>> On 06/05/21 08:23AM, Simon Glass wrote:
>>> Add a function to duplicate a memory region, a little like strdup().
>>>
>>> Signed-off-by: Simon Glass <sjg at chromium.org>
>>> ---
>>>
>>> Changes in v2:
>>> - Add a patch to introduce a memdup() function
>>>
>>> include/linux/string.h | 13 +++++++++++++
>>> lib/string.c | 13 +++++++++++++
>>> test/lib/string.c | 32 ++++++++++++++++++++++++++++++++
>>> 3 files changed, 58 insertions(+)
>>>
>>> diff --git a/include/linux/string.h b/include/linux/string.h
>>> index dd255f21633..3169c93796e 100644
>>> --- a/include/linux/string.h
>>> +++ b/include/linux/string.h
>>> @@ -129,6 +129,19 @@ extern void * memchr(const void *,int,__kernel_size_t);
>>> void *memchr_inv(const void *, int, size_t);
>>> #endif
>>>
>>> +/**
>>> + * memdup() - allocate a buffer and copy in the contents
>>> + *
>>> + * Note that this returns a valid pointer even if @len is 0
>>
>> I'm uneducated about U-Boot's memory allocator. But I wonder how it
>> returns a valid pointer even on 0 length allocations. What location does
>> it point to? What are users expected to do with that pointer? They
>> obviously can't read/write to it since it is supposed to be a 0 byte
>> long allocation. If another positive length allocation happens before
>> the said pointer is freed, will it point to the same memory location? If
>> not, isn't the 0-length pointer actually at least a 1-length pointer?
>
> I think it is just a 0-length pointer and that the only thing you can
> do with it is call free().
>
> I am certainly no expert on this sort of thing though. It seems that
> some implementations return NULL for a zero size, some return a valid
> pointer which can be passed to free(). Of course, U-Boot lets you pass
> NULL to free() anyway.
dlmalloc has a minimum allocation size of 2*sizeof(void *) (e.g.
MINSIZE - 2*SIZE_SZ). If you request less than this, you will get an
allocation of this size. This is the same as other requests, which are
rounded up the the nearest multiple of MALLOC_ALIGNMENT. Of course,
malloc_simple will actually give you a zero-byte allocation, so don't
rely on being able to store anything there ;)
--Sean
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