Il 14/03/2018 20:26, goldsi...@gmx.de ha scritto:
On 14.03.2018 16:42, Giuseppe Modugno wrote:
I was experimenting with lwip memory allocator. I defined:
#define MEM_LIBC_MALLOC 0
#define MEM_USE_POOLS 0
#define MEM_SIZE (32 * 1024)
#define LWIP_RAM_HEAP_POINTER ( (void *)0x2007C000 )
I'm using LPC1768 that has a 32kB SRAM block starting from address
0x2007C000. However I have a Hard Fault error during mem_init(), because
ram_end is initialized to 0x2007C000 + (32*1024), that is over the
available memory.
So MEM_SIZE should be smaller than available heap memory.
Is that a question? Or an action request?
It was a question, even if I forgot to add a question mark :-)
To my understanding, the comment above #ifndef LWIP_RAM_HEAP_POINTER
in mem.c clearly states that enough memory is required. It does NOT
say that amount is MEM_SIZE. If you want, create a patch with a better
documentation and we could apply it.
Ah ok, I now understand the comment "we need one struct mem at the end
and some room for alignment".
MEM_SIZE isn't the total available memory for heap management (including
accessory structs), but it is the *useful* memory space the application
needs for the heap (excluding accessory structs). For example, if
MEM_SIZE is 1024, I can be sure one malloc(1024) won't fail.
In my case, I have a completely free (not used by the linker) 32kB RAM
region starting from 0x2007C000. To instruct lwip to use that region for
heap, I think I have to define:
#define MEM_SIZE (32 * 1024 - 1 * SIZEOF_STRUCT_MEM)
#define LWIP_RAM_HEAP_POINTER ( (void *)0x2007C000 )
Supposing we don't need other additional space for alignment.
In this case the total free heap memory available for data is exactly
MEM_SIZE, that is less than 32kB.
Is it correct?
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