alt Hacker Newsbut without virtual memory you can't move them around and so program that wants to allocate 2MB of 64k segments can't run if there is no 2MB continuos hole
but without virtual memory you can't move them around and so program that wants to allocate 2MB of 64k segments can't run if there is no 2MB continuos hole
I should preface this by saying I'm taking about x86 segmentation in general. On the 8086 you're right, but the 8086 can't address 2MB of memory to begin with. On the 80286 in protected mode, the situation is different in the way I'm about to describe.
The memory itself doesn't have to be contiguous.
2MB of 64K segments maps to 32 segments. So you need 32 locations in physical memory capable of storing 64K.
The programming model for addressing that block of memory necessarily includes both segment selectors and offsets. The segment selectors are indices into a segment table that contains the base address of each of the 32 segments. As long as the segment selectors themselves can be allocated contiguously in the segment table, you have enough to be able to compute which segment you need for which address in the 2MB range. It's the indirection through the segments table that maps it to physical addresses that do not need to be contiguous.
Raymond Chen talks a bit about how it worked in Windows 3.x here: https://devblogs.microsoft.com/oldnewthing/20171113-00/?p=97...