[senfiaj]

For its time it was a decent idea. Software was smaller and simpler. But today (and even before 64-bit) software is larger, more complex, we also need memory protection / isolation and more flexible memory allocation / sharing, so paging memory was not introduced for nothing.

[flohofwoe]

> we also need memory protection / isolation

I seem to remember that memory segments came with a permission system (read-only, read/write, execute) in 'protected mode'. Probably only added in the 286 though (I was always more of an m68k guy at that time).

[senfiaj]

Maybe (I think it's possible in protected mode), but it still has an allocation problem, imagine there are programs A, B, C in the memory. Later, A and C are unloaded, leaving 2 free holes, totaling in 2MB. Now you want to load a 2MB program, but there is no unfragmented 2MB free block. The only solution I see, is to shift some loaded programs, which might be slow and even risky. Paging makes this problem much easier. Also, paging makes permissions and memory sharing more granular.

[mschaef]

There is no such thing as a "2MB program".... all you have is a program composed of <=64K segments, which are easy enough to fit into the hole.

If you do need something approaching a 2MB block of memory, you don't need a contiguous range of memory, what you need is a contiguous range of selectors, which is a different (and probably easier) problem to solve.

[PunchyHamster]

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

[mschaef]

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...

[senfiaj]

>> 8086 can't address 2MB of memory

This was just for illustration, not claiming that actual 8086 does this.

>> Raymond Chen talks a bit about how it worked in Windows 3.x here: https://devblogs.microsoft.com/oldnewthing/20171113-00/?p=97...

And this is the problem, it was very painful just to walk through a 200 KB buffer. This required compiler/runtime tricks, different selector increments in real vs protected mode, and special pointer types. Paging later made this kind of thing look like one flat array, a thing segmentation could not: making non-contiguous physical RAM appear contiguous to the program.

[pjc50]

I was under the impression that the permissions only kicked in once you were in 32-bit mode on the 386, what Windows called "386 Enhanced" mode.