An opcode is an instruction. An operand is information used by the opcode. Not all opcodes require operands.

The microprocessor uses an opcode fetch cycle for every instruction because it has to know the opcode in order to execute it, and that is located in memory.

3 for opcode fetch, 1 for opcode decode, 3 for operand fetch, and 3 for opcode store, for a total of 10, not including wait states.

Hi, The equation to find the number of instructions with n-bit opcode is 2^n. If your opcode is n=4, the it's 2^4 which is 16. So with a four bit opcode you can have 16 different instructions. ---- 11 levels

The process of transferring instruction codes from memory location to instruction queue register is called opcode fetch.

mov H , L mov is opcode H L are operands

The instruction opcode is a type of data contained in memory, pointed to by the PC (Program Counter) register.

popfd

IP is incremented after fetch of instruction opcode. Specifically, IP is incremented by the number of opcode bytes.

Each mnemonic maps directly to a machine instruction code, known as an opcode. Some mnemonics map to more than one opcode, however the instruction's operand types will determine which specific opcode will be generated.

As far as the bus interface is concerned, there is no real difference between data and instructions. Except for the S0 pin, an opcode fetch will look the same as a memory read. There is one extra clock cycle following an opcode fetch, which is used by the CPU to decode and process the opcode, but the bus does not care because there is no sequence initiation with ALE.

The hard way: Download the processor manuals and code the opcode and operands by hand

The easy way: Use an assembler program. The instructions are slightly different for each program, so try reading the manuals.

the opcode is fetched from the memory and decoded

i) Instruction code deals only with mnemonics and its corresponding opcode but data code refers to your data like 10h which is always of 8 bits or a particular address say 8080h which is of 16 bits.

ii) Data is your input to the instruction but an opcode is native to your machine.

iii) Data is user specific instruction while opcode is machine specific instruction

iv) You can alter data code but you cannot modify an instruction opcode.

In the 8085, the LDA instruction loads the accumulator from memory, while the STA instructionstores the accumulator to memory. LDA is a read, while STA is a write. LDA is opcode 3AH, while STA is opcode 32H.

In the 8085, the LDA instruction loads the accumulator from memory, while the STA instruction stores the accumulator to memory. LDA is a read, while STA is a write. LDA is opcode 3AH, while STA is opcode 32H.

three

RST is simply the opcode chosen to represent the Restart instruction.

LDA is an Intel 8085 opcode, 3AH, that loads that accumulator from a location specified in memory.

The OUT instruction on the 8085 uses 10 T cycles, 3 for opcode fetch, 1 for opcode decode, 3 for port address fetch, and 3 for port data store. Any wait states encountered are above and beyond that.

The IR is a register constisting of two parts, opcode and address. The opcode is decoded and gives instructions to the control unit whilst the address part is used to give the location in the memory of for example data required for a certain process

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.

op code is used as the value of instruction . And operand is address location where the instruction can meet.

That's what opcode 'NEG' does on some processors.Can be emulated with a 'NOT' and an 'INC'.

no.it is not necessary because one can make timing 010

By the opcode. E9 - JMP Direct, EB - JMP Short

Yes and no. INTR response requires an opcode fetch sequence from the interrupting hardware device, often a CALL instruction, so there is no vector table in memory for it, because you can CALL any location. On the other hand, some implementations provide an RST instruction as the opcode, making it a vectored interrupt.

That's already been done. Its called a DEC PDP-8.

A microprocessor know whether the next byte is an instruction or data because the microprocessor knows for what it is looking. The bus, on the other hand, for an 8085 based system, knows an opcode from data by looking at S0 and S1 when IO/M- is low. If both are high, it is an opcode, otherwise it is data.

from what ive seen yes. Centsports is a free online sports betting site: http://www.centsports.com/?opcode=351728

Three for opcode fetch, one for decode, two to process the manipulation of the stack pointer.

A code that leaves a spare bit to indicate that if that bit is set, consider this byte and the next byte to be defining the entire code.

The 8086 microprocessor differentiates between opcodes and instruction data through its instruction format, where the first byte(s) typically represent the opcode, while subsequent bytes represent operands or data. The opcode specifies the operation to be performed, while the data can include registers, memory addresses, or constants. Additionally, the instruction pointer and segment registers help the processor understand the context of the instruction, allowing it to interpret the opcode and associated data correctly. This structured format enables the 8086 to efficiently decode and execute instructions.

Immediate addressing mode is when one of the operands is "immediately" located after the opcode. It is more correct to say that the operand is part of the instruction.

Immediate addressing mode is when one of the operands is "immediately" located after the opcode. It is more correct to say that the operand is part of the instruction.

The STA 4200H instruction in the 8085 requires 4 machine cycles and 13 T states to complete its fetch, processing, and execution.

Cycle One: Opcode fetch, 3 T states plus one opcode process state.

Cycle Two: Opcode address byte 00H fetch, 3 T states

Cycle Three: Opcode address byte 42H fetch, 3 T states

Cycle Four: Accumulator store, 3 T states.

Each cycle will have additional T-Ready states as needed by the READY pin. 13 T states is the minimum.

The LDA instruction will also require 13 T states, with the last cycle being a read cycle instead of a write cycle.

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.

It is partial with opcode and processor code. It is meaningless when we see it on our eyes, but it provides info instructions to Java Virtual Machine to execute that.

An instruction only has one opcode.

There may be two opcode bytes, or there may be different fields in the assembly code of the instruction, but the bit pattern of an instruction will always generate the same results.

There are some results that can be generated with two different opcodes. SUB A and XRA A, for instance, both clear the accumulator, but they are two different opcodes for two different instructions.

Log2 260 is 8.022, so it would seem that 9 bits are required to handle 260 instructions. In practice, however, the opcode is a multiple of 8 bits, so most instructions are 8 bits, with a few being 16 bits.

The part of the instruction that tells a computer what operation to perform is variously called the "operation code", "op code", "opcode", "operation", "order code", "instruction code", "function designator", "function", "prefix", "designator", etc. depending on the specific computer and the (arbitrary) preferences of the designers of that computer architecture about terminology.

Some computers have special instructions that use parts of the instruction in a different way than other instructions do to provide additional operations (e.g. PDP-8 in the OPR instruction used the fields used by most other instructions for memory addressing as a "microcoded" operation request, PowerPC has "primary opcode" and an optional "extended opcode").

There are some computer architectures that do not even use instructions (e.g. dataflow computers) or have instructions without an opcode (e.g. Transport Triggered Architectures, Forth virtual machine) but they are still fairly rare.

The assembler's role is most important.it converts the high level language statements into machine level language statements with the help of some operand and opcode specifications.there is first mnemonic opcode specification.here instead of writing binary opcodes,mnemonic opcodes can be specified.advantage of using mnemonic opcodes are:program becomes readable.debugging becomes simple.so it is the responsibility of the assembler to replace each mnemonic opcode by its respective binary opcode.also there is symbolic operand specification.in that,instead of specifying the addresses of instructions and data,symbols can be used.advantage of using the symbolic operand is that the program can be modified with no overhead.it is the responsibility of assembler to replace each symbol by its address.

Every instruction contains to parts: operation code[opcode],and operand.

The first part of an instruction which specifies the task to be performed by the computer is called opcode.

The second part of the instruction is the data to be operated on.,and it is called operand. The operand[or data]given in the instruction may be in various forms such as 8-bit or 16-bit data, 8-bit or 16-bit address, internal register or a register or memory location.

It is a 3 byte instruction, with one byte for opcode and the other two for the 16bit address. It takes four machine cycles (one to fetch opcode, one to fetch lower order address, one to fetch higher order address and another one to fetch the data from the memory)... i.e. it takes 13 time states to perform the LDA instruction

different microprocessors take different number of states. without knowing processor its not possible to comment.