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https://www.righto.com/2020/07/the-intel-8086-processors-registers.html
The photo shows the silicon die of the 8086 processor under a microscope. The metal layer on top of the chip is visible, with the silicon hidden underneath. Around the outside edge, bond wires connect pads on the die to the chip's 40 external pins.
The highlighted region indicates the 8086's fifteen 16-bit registers and six bytes of instruction prefetch queue.1 Registers take up a significant portion of the die, even though they are just 36 bytes in total. Due to space limitations, early microprocessors had a relatively small number of registers; in comparison, a modern processor chip has kilobytes of registers and megabytes of cache storage.2
[...] The 8086 and other chips of that era were built from a type of transistor called NMOS. These chips consisted of a silicon substrate, which was "doped" by diffusion of arsenic or boron to form transistors. Above the silicon, polysilicon wiring created the gates of the transistors and wired components together. Finally, a metal layer on top provided more wiring. (Modern processors, in comparison, use CMOS technology, which combines NMOS and PMOS transistors, and they have many metal layers.)
(Score: 3, Interesting) by owl on Thursday April 27, @10:54PM (1 child)
For the 8086, the answer is no. The registers are just storage, and any 'operation' on them means their contents is sent to the ALU, the ALU performs the requested "operation" and the result from the ALU overwrites the destination register's old contents.
There are designs similar to what you are thinking of, where the "operation logic" is embedded close to the memory where the data is stored, but the 8086 is not an example of that type of design.
(Score: 2) by DannyB on Friday April 28, @03:43PM
Thanks
How often should I have my memory checked? I used to know but...