Re: [lwip-users] MEM_SIZE

[Giuseppe Modugno]

Il 14/03/2018 20:26, address@hidden 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?