maloc enhancements

Improvements need to be make when working with the avl5 list as it deals with fragmentation . Plus additional data collection reports need to be made available to better detect and analysis the issue.
In this case there was an application requesting a 4KB maloc . There was roughly 2.8MB of storage available all in 3k pieces, but none available for the 4k request. .

This is what the dump showed:
Looking at heap buffers in use. There were numerous gaps in heap addresses which indicate available storage but those gaps were all about 3K (x'C08' was most common gap). I found over 950 of these 3K available storage chunks which is about 2.8MB of storage...none of which can be used for a 4KB malloc request.

This fragmentation seems to show up as a pattern:

1.Ask for 4095 or 4096 bytes of heap storage via malloc/calloc/raloc that ends up requiring 4104 bytes (4096+8 byte trailer used for heap corruption detection).

2.4104 bytes is greater than AVL4 size (4096) so request is handled from AVL5 list,
a.If there is nothing on AVL5, a new 1MB frames is allocated to back another MB of heap storage
and 4104 byte chunk is taken from the MB.
b.If there is a large enough AVL5 item, 4104 byte chunk is taken from this item.

3.Allocation is released going back on AVL5 list as 4104 byte item.

4.New request for x'200' or x'348' of heap storage. This request is bumped up to x'400' (AVL3 size = 1024). If there is nothing on AVL3 list, then AVL5 list is checked.

5.The AVL5 list has the 4104 byte allocation to use for this request. AVL5 processing attempts to split the 4104 byte allocation into 2 pieces: the requested 1024 byte piece and remainder (3080 or x'C08' bytes) to be put back on the AVL5 list.

6.However, AVL5 list processing will only break the allocation into 2 pieces if the remainder is large enough to be another AVL5 item. This means it must be greater in size than AVL4 items (4096 bytes). So the entire 4104 byte allocation is returned as the storage for the request for x'200' or x'348' bytes.

7.Ask for another 4095 or 4096 byte allocation and repeat above process....keep using 4104 bytes for x'200' and x'348' byte allocations, depleting the heap of storage.