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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.
There are 74 instructions in the 8085 microprocessor.
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.
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.
Copied from discussion, as no response has been received from the discussion... There is no branch or jump instruction in the 8085 or 8086/8088 that has a 10 bit address field. Please specify what microprocessor to which you are referring.
first connect the 8255 to the 8086 microprocessor and configure it using the CWR then connect the key board to the 8255
There are many instructions in the 8085. To find the hex code for a particular instruction, you need to look at the data sheet. For more information, please see the Related Link below.
The need for a timing diagram for a microprocessor is, primarily, to allow you to properly design the interface logic that will support the microprocessor. You need to know what lines are used to execute various data transfers, what are the timing of those lines with respect to each other, and how the microprocessor expects the external logic to behave. You can also use the timing diagram to understand how the microprocessor functions and, particularly, to know how long each instruction will take.
53 bytes long
The sizeof long int is platform-dependent, often 4 bytes or 8 bytes.
It depends on whether the machine code is one, two, or three bytes long, and on whether or not the instruction transferred control to another location. In the case of a non-jump single byte instruction, the PC will have a value of 2060H after the instruction is complete, and it will be 2061H or 2062H after a two or three byte instruction. In the case of a jump, call, or interrupt, the PC will depend on the instruction.
A Mac address is a 48bit addressing scheme (usually represented in HEX). There are 8 bits in a bytes therefore it is 6 bytes long.