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