In Depth: Process Heap Layout
An Erlang process has a young heap (where all new data is placed), old heap (possibly this one does not exist, contains data which survived previous garbage collection) and a stack, which is located inside young heap.
Also, a process has access to read-only literal area, which belongs to one or more of Erlang modules, it contains constant literals which are found in module code. At some rare occasions (code upgrade or purge) literal area can be copied to old heap by means of garbage-collection (see erts_garbage_collect_literals in erl_gc.c).
A process has chain of Off-heap objects (so-called MSO -- mark and sweep object list). The head of the chain is a field in Process struct, and the chain continues through the heap visiting Header Tag objects such as refc binaries and fun closures. During garbage collection this chain is followed and actions can be taken (similar to C++ destructors) when an MSO object is going to die -- for example refc binaries have their refcount reduced and possibly are freed. MSO objects are strictly sorted from new to old, by the order of their creation on heap.
How Heap Grows
Additionally, heap segments can exist (for situations, when heap is not enough, but data must be allocated). Heap segments are consolidated before garbage collection and merged onto heap.
Heap top (heap_top) marks current allocation position in heap. When more data is needed, heap_top is advanced forward and data goes there.
Heap always grows forward from heap_start (see erl_process.h, Process struct, fields heap, old_heap). Stack always grows backwards from heap_end to heap_top, as soon as stack meets heap top, the young heap is considered full and garbage collection is triggered.
Old heap does not contain a stack. Old heap can only grow during garbage collection and its future size is precalculated.
To minimize memory fragmentation, heap sizes like a puzzle pieces are allocated following Fibonacci sequence (starting from 12 and 38 Words) for 23 total steps, after which (at about 1.3 million Words) sequence continues by adding 20% to the previous value.