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When each address in memory(RAM) stores 8bits of data. 8086 have access to 2^20 address locations making it as (2^20)*8 = 1MB. When it has 16bit data bus and accessing only 8bit data then what is the use of 16bit data bus.
Q1) Does it access two address locations and increment IP(instruction pointer) by two.
Q2) Can it use two banks of memory simultaneously and increment IP by one as shown in below image.
Is it true if not why
The memory is byte addressable, and the 16-bit bus transfers two bytes at a time.
But the total capacity is still just 1 MB.
The diagram is a bit misleading in this regard — it should show only A19-A1 going to both banks. A "bank" refers to 8-bit wide memory, and one or both banks are accessed as described in the table. They're trying to show that the left bank (D15-D8) is enabled only when BHE- is low, and the right bank (D7-D0) is enabled only when A0 is low.
So yes, when stepping through memory (such as when prefetching instructions), the bus address increments by two.
However, note that 8086 instructions are anywhere from 1 to 6 bytes long, and that's what determines the actual IP increment. The 8086 reads instructions from a 6-byte prefetch buffer (4 bytes on the 8088), and the logic that fills that buffer uses its own address register that is distinct from the IP.
mov word [si + 1234], 1234). Later CPUs impose a 15 byte limit, otherwise taking a #UD undefined instruction exception.
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Nov 10, 2019 at 15:14
cs: segment with prefixes, 8086 will wrap around IP and keep trying to decode forever, never reaching the end of this "infinite" instruction.
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Nov 10, 2019 at 15:22