[PATCH v2 01/50] lib: Add memdup()

[Simon Glass]

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.

Regards,
Simon