3 ms·

> a 32-bit OS can only address 3.2 GB of RAM I am wondering how these numbers stick together. I especially wonder why those extra 0.8 GB are unaccessible, when

• Netch 3y ago
  The design of current x86s is that multiple regions under 4GB boundary are occupied by I/O regions, some of them tied to processor (as APIC, HPET, etc.) and some to extension cards which reject working in 64-bit mode. 3.2GB is an example figure - it could be 3.5GB, 3.6GB... - but reflects that RAM can't be mapped at all of lower 4GB. Memory controllers typically remap part of it to a range upper than 4GB boundary, so 32-bit PAE-capable or 64-bit OS can still use it. OTOH using 64-bit OS is recommended with RAM >=1.5GB because if OS can't map all RAM _twice_ and all device memory to virtual address space, tricks with memory banking (frequent page remapping) are getting needed to deal with data move.
• pvillano 3y ago
  Invoking Cunningham's, I'd guess it's memory mapped I/O occupying 0xE0000000 - 0xFFFFFFFF. That's the top ⅛ of the address space, which is every binary address starting with 111. In researching this I learned that some 32bit hardware could utilize 128GB of ram using Physical Address Extensions. Each process is still limited to 4GB though.