"When I set MEM_ALIGNMENT to 4, why sometimes mem_malloc() returns a 4
byte aligned address and sometimes it returns a 8 byte address?" Well
when set to 4 Byte alignment, you will likely have a 50/50 chance of the
address being 8 Byte aligned (depends on the size of items you are
allocating, however it will always be 4 Byte aligned).
Terry
On 27/05/2019 13:55, [email protected] wrote:
Hi Terry.
My cpu is a processor based on sparc v8 architecture. I have checked
the sparc v8 manual and leared that the processor uses LDD instruction
to load double word (64 bit) and it requires the address is double
word (64 bit) aligned.
However, in another application when I use mem_malloc() to allocate
memory for structure interface, the returned memory pointer is
actually 8 byte aligned.
So, I am wondering whether there is a memory alignment assumption for
mem_malloc().
When I set MEM_ALIGNMENT to 4, why sometimes mem_malloc() returns a 4
byte aligned address and sometimes it returns a 8 byte address?
Nevertheless, your suggestion to set MEM_ALIGNMENT to 8 seems to be a
good solution to this problem.
yan
On 05/27/2019 16:18, Terry Barnaby <mailto:[email protected]> wrote:
But what is your CPU type ?
Some CPU's can load/save an 8 Byte (64 bit) value (such as
uint64_t, double) on a 4 Byte boundary, while some others cannot.
I suspect you need to set MEM_ALIGNMENT to 8 for your CPU.
Terry
On 27/05/2019 06:49, [email protected] wrote:
Hi, Terry. Thanks for your reply.
The memory alignment I set in lwipopts.h is 4 as follows. Since I
am using a 32 bit processor.
#define MEM_ALIGNMENT 4
On 05/27/2019 13:16, Terry Barnaby
<mailto:[email protected]> wrote:
In your lwipopts.h what have you set for MEM_ALIGNMENT ?
// MEM_ALIGNMENT: should be set to the alignment of the CPU
for which lwIP is compiled.
#define MEM_ALIGNMENT 4
Terry
On 27/05/2019 04:17, [email protected] wrote:
Could anyone give me some advice？
I recently use mem_alloc() to allocate a structure memory
which have a int64 member. However, mem_alloc() returns a
memory poniter which is 4 byte aligned which cause a memory
unaligned fault.
The code is as follows,
ifp=mem_alloc(sizeof(structure interface));
The structure intrface has a member flags which is int64.
The size of structure interface is 112 which is a multiple
of 8 while the address returned is 0x720ef324 which is not 8
byte aligned. So when I want to access ifp->flags, an
unaligned fault will be triggered.
Could any give some hint?
Thanks!
Best regards,
yan
On 05/25/2019 22:19, yanhc519
<mailto:[email protected]> wrote:
Hi all.
What is the memory alignment of the pointer returned by
mem_alloc() in mem.c?
Since the only argument of mem_alloc() is size, so what
alignment will mem_alloc() choose?
In http://man7.org/linux/man-pages/man3/malloc.3.html,
it says "/The malloc() and calloc() functions return a
pointer to the allocated memory, which is suitably
aligned for any built-in type./"
Since the largest type is double or int64 which are 8
bytes, so I guess /malloc() /in linux will choose
alignment of 8 bytes.
Is this guess also applied to mem_alloc() in LwIP?
That is, does mem_alloc() in LwIP also choose alignment
of 8 bytes?
Thanks!
Best regard,
yan
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