[MagerValp]

But segmented memory isn’t a requirement, that technique works just as well with linear address space. You use a single 32-bit base pointer and then store 16-bit offsets for your data. We used that all the time on 68K and other architectures.

[skissane]

The difference is that 68K is designed as fundamentally a 32-bit architecture. (Even though the original implementation was physically 16-bit.)

Whereas 8086 is a 16-bit architecture with an extended address space.

The use of segmentation to enable a 16-bit architecture to address more than 64K was not original to the 8086, many 16-bit minicomputers (e.g. the PDP-11) used the same basic idea, although the specific implementation Intel chose was rather unique

Part of why the 8086 was 16-bit not 32-bit, was to make it easier to port software to it from the 8080, which was an 8-bit architecture with 16-bit addressing. It also was likely one of the reasons why the 8086/8088 was cheaper than the 68K, which is part of why IBM chose it over the 68K for the IBM PC

[p_l]

I heard opinions that 8086 won for IBM PC project because it was available for longer and more proven, with more options for second-sourcing.

[actionfromafar]

That would be fact - AMD was the second source for the 8088. (Licensing IP from Intel.)

[p_l]

Motorola 68000 also had second sourcing options, just like 8088 - but 8086 was released a year earlier and built up on well-known successful design of 8080, even if they weren't exactly compatible. It built confidence about Intel's ability to deliver, even when 8088 was released the same year as 68k, as it was just bus-narrowed cheaper version, not an entire architecture.

[actionfromafar]

Which reminds me - the bus-narrowed version was important, because they then could use cheap and off the shelf 8-bit components for the rest of the chip set.

[p_l]

Both required a bus demultiplexer, but indeed 8088 was cheaper in a bunch of places (both 8086 and 8088 could run 8bit data transfers, though)