32 byte
no
6 bytes
In 8086 the instruction queue is 6 byte long. This is because even the longest 8086 instruction is 6 byte long. Thus it is possible to prefetch even the longest instruction in the instruction set.
it goes to queue for next instruction
32 byte
6 bytes
increases speed
An instruction queue is used in the 8086 to speed up the average time it takes to process an instruction. Some instructions are faster than the bus, while some are slower. If the CPU had to wait for all of the instructions, there would be gaps of time where the CPU is doing nothing. The queue helps to eliminate that gap by prefetching instructions in the hope that they will be ready for use when the CPU gets to them.
Because that's how Intel designed it. They chose a 6 byte queue in the 8086 in order to optimize speed versus latency of the execution unit versus the bus interface unit. The decision for 6 bytes, as compared to 4 or 8 or some other number was a cost tradeoff and a recognition of the average mix of instruction execution cycle times in a typical processing thread.
pata nai
The process of transferring instruction codes from memory location to instruction queue register is called opcode fetch.
In the context of the 8086 microprocessor, the prefetch queue is a 6-byte high-speed queue that stores the next instruction to be executed. This helps improve performance by fetching instructions in advance and storing them in the queue. When the CPU is ready to execute the next instruction, it can quickly access it from the prefetch queue without having to wait for the instruction to be fetched from memory.