32 byte
no
The 8086/8088 instruction queue is a buffer that holds opcode bytes that have been prefetched by the bus interface unit. This speed up operations of the processor by helping to reduce fetches latency, i.e. to improve the probability that an opcode byte fetched by the processor is already available. This works best when there is no branching, as a branch would invalidate the queue. Advanced processors attempt to "predict" the branch, making the probability even better.
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.
The 8086/8088 instruction queue is a buffer that holds opcode bytes that have been prefetched by the bus interface unit. This speeds up operations of the processor by helping to reduce fetch latency, i.e. to improve the probability that an opcode byte fetched by the processor is already available.
it goes to queue for next instruction
32 byte
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.
6 bytes
6 bytes
increases speed
The process of transferring instruction codes from memory location to instruction queue register is called opcode fetch.
There is no PC register in the 8086/8088. It is called the IP register by Intel and it stands for the Instruction Pointer. It contains the address of the current/next instruction to be executed.