Thanks Terry! On 05/27/2019 21:04, Terry Barnaby wrote: "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 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
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 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
_______________________________________________
lwip-users mailing list
[email protected]
https://lists.nongnu.org/mailman/listinfo/lwip-users
_______________________________________________
lwip-users mailing list
[email protected]
https://lists.nongnu.org/mailman/listinfo/lwip-users
_______________________________________________
lwip-users mailing list
[email protected]
https://lists.nongnu.org/mailman/listinfo/lwip-users
_______________________________________________
lwip-users mailing list
[email protected]
https://lists.nongnu.org/mailman/listinfo/lwip-users
