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
The length of the instruction queue in the 8086 microprocessor is limited primarily to manage complexity and ensure efficient execution. A shorter queue reduces the risk of delays caused by branch instructions, which may invalidate queued instructions. Additionally, limiting the queue size helps maintain control over the instruction fetch cycle, ensuring that the CPU can quickly process instructions while minimizing the overhead associated with managing a larger queue. Overall, this design choice balances performance with the microprocessor's architectural simplicity.
6 bytes
32 byte
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.