4 ms·

> Because Linux-capable application processors have a memory management unit, *alloc() calls execute swiftly and reliably. Physical memory is only reserved (fau

> Because Linux-capable application processors have a memory management unit, *alloc() calls execute swiftly and reliably. Physical memory is only reserved (faulted in) when you actually access a memory location. Memory fragmentation is much less an issue since Linux frees and reorganizes pages behind the scenes. This paragraph is confusing to me. - How would a MMU help for malloc to execute "swiftly and reliably" ? - What does "actually access a memory location" implies exactly? - "since Linux frees and reorganizes pages behind the scenes": Is this not what people dislike from malloc? That it's none deterministic?

spatular 6y ago
MMU allows linux to provide a "flat" memory space to each userspace process. This memory space is assembled from 4KB pages that can be randomly dispersed throughout physical memory. Say, you want to malloc() 1MB of memory on a system that's been up for some time. Physical memory may not have a fitting continuos chunk of memory, but virtual memory of a newly created process will always have one, provided that there is enough free pages available. In other words without MMU all programs share the same memory space, and if it get fragmented, generally you'd have to reboot. With MMU fragmentation can still be an issue, but it's greatly reduced, since each process has its own memory map. And if memory of the process gets framgmented, you can just restart it. If runtime supports object compaction/relocation, then fragmentation may be not an issue at all. Re access to memory locations -- mostly any direct access to instruction or data memory in userspace program automatically goes through MMU remapping.
- vitaminCPP 6y ago
  Thanks. I realized that there are several types of MMU. I was thinking of the simple MMU (on 32bit MCU) that limit memory access. Not the one that provides the illusion of virtual addressing.
  - spatular 6y ago
    To be precise, that'd be called MPU -- memory protection unit.

spatular 6y ago

MMU allows linux to provide a "flat" memory space to each userspace process. This memory space is assembled from 4KB pages that can be randomly dispersed throughout physical memory. Say, you want to malloc() 1MB of memory on a system that's been up for some time. Physical memory may not have a fitting continuos chunk of memory, but virtual memory of a newly created process will always have one, provided that there is enough free pages available. In other words without MMU all programs share the same memory space, and if it get fragmented, generally you'd have to reboot. With MMU fragmentation can still be an issue, but it's greatly reduced, since each process has its own memory map. And if memory of the process gets framgmented, you can just restart it. If runtime supports object compaction/relocation, then fragmentation may be not an issue at all. Re access to memory locations -- mostly any direct access to instruction or data memory in userspace program automatically goes through MMU remapping.

vitaminCPP 6y ago
Thanks. I realized that there are several types of MMU. I was thinking of the simple MMU (on 32bit MCU) that limit memory access. Not the one that provides the illusion of virtual addressing.
- spatular 6y ago
  To be precise, that'd be called MPU -- memory protection unit.