Hi Pratyush, On Thu, 6 May 2021 at 10:07, Pratyush Yadav <p.ya...@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 <s...@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
