The only reason why it is tough for us to use nano malloc
is because of the small shortcoming where nano_malloc()
splits a bigger chunk from the free list into two pieces
while handing back the second one (the tail) to the user.
This is error prone and especially bad for smaller heaps,
where nano malloc is supposed to be superior. The normal
malloc doesn't have this issue and we need to use it even
though it costs us ~2k bytes compared to nano-malloc.
The problem arise especially after giving back _every_
malloced memory to the heap and then starting to exercise
the heap again by allocating something small. This small
item might split the whole heap in two equally big parts
depending on how the heap has been exercised before.
I have uploaded the smallest possible application
(only tested on ST and Nordic devices) to show the issue
while the real customer applications are far more complicated:
https://drive.google.com/file/d/1kfSC2KOm3Os3mI7EBd-U0j63qVs8xMbt/view?usp=sharing
The application works like the following pseudo code,
where we assume a heap of 100 bytes
(I haven't taken padding and other nitty and gritty
details into account. Everything to simplify understanding):
void *ptr = malloc(52); // We get 52 bytes and we have
// 48 bytes to use.
free(ptr); // Hand back the 52 bytes to nano_malloc
// This is the magic line that shows the issue of
// nano_malloc
ptr = malloc(1); // Nano malloc will split the 52 bytes
// in the free list and hand you a pointer
// somewhere in the
// middle of the heap.
ptr2 = malloc(52); // Out of memory...
I have done a fix which hands back the first part of the
splitted chunk. Once this is fixed we obviously
have the 1 byte placed in position 0 of the heap instead
of somewhere in the middle.
However, this won't let us malloc 52 new bytes even though
we potentially have 99 bytes left to use in the heap. The
reason is that when we try to do the allocation,
nano-malloc looks into the free list and sees a 51 byte
chunk to be used.
This is not big enough so nano-malloc decides to call
sbrk for _another_ 52 bytes which is not possible since
there is only 48 bytes left to ask for.
The solution for this problem is to check if the last
item in the free list is adjacent to sbrk(0). If it is,
as it is in this case, we can just ask sbrk for the
remainder of what is needed. In this case 1 byte.
NB! I have only tested the solution on our ST device.
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