The following is an 8051 instruction: CJNE A,#'Q',AHEAD a. What is the opcode for this instruction? b. How many bytes long is this instruction? c. Explain the purpose of each byte of this instruction. d. How many machine cycles are required to execute this instruction? e. If an 8051 is operating from a 10 MHz crystal, how long does this instruction take to execute?

Please answer all the following questions. 1. In a hypothetical computer, the processor has four registers: an 8-bit Program Counter (PC), a 16-bit Accumulator (AC), a 16-bit Counter (CTR), and a 12-bit Pointer (PTR). The memory is divided into words each of which is 16-bit long. Each word can hold either an instruction or a piece of data. For each instruction X, the four most significant bits (denoted by X15-12) represent an opcode. The rest of the instruction (denoted by X11-0) can be either an address or a value of an operand. The table below explains some of the instructions supported by the processor. Opcode (binary) 0000 0001 0010 0011 0100 0101 0110 0111 1000 1001 1010 1011 Operation Load AC from a memory location whose address is X11-0. Store AC into a memory location whose address is X11-0. Add to AC the contents of a memory location whose address is the value of PTR. The result is saved to AC. 1100 Subtract from AC the contents of a memory location whose address is the value…

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what are the bits in s1 in 8 hexdecimal digits after the execution.

7. For both of the instructions shown below, the initial conditions are shown. The result (which the instruction stores in r1) is also shown. For each instruction, answer these questions: • What is the state of the flags (NZCV) after the instruction executes • If the operands were signed, are the results valid? Why or why not? • If the operands were unsigned, are the results valid? Why or why not? a) initial conditions: r0 = 0xAAAA_AAAA rl=0x5555_5555 ADDS rl,rl,ro Result: r1=0xFFFF_FFFF b) initial conditions: r0=0x7FFF_FFFD r1 = 0x0000_000F ADDS rl,ro,r1 Result: r1=0x8000_000C

Given the number Ox21100006, what is it when interpreted as a MIPS instruction? 1. The instruction format is type |. (R, I, or J?) 2. The opcode value is (specify as a hexadecimal value, e.g., OxfEE3, lower-case x, upper case hex values.)

1.Address of the opcode; 5CD   2.Address of the operand for the operation; 5DE   3.None of these   4.Address of the operand for the operation; 5CD

What is the reverse assembly of OxE8071020 machine language instruction? The Op-Code of the instruction is: The operands of the instruction are:

0 Suppose we have the instruction Load 800. Given memory as follows (assume all numbers are decimal): Memory 600 1200 800 What would be loaded into the AC if the addressing mode for the operation is direct addressing? Question 21 800 900 1000 1100 900 500

Because of this, the means by which one instruction acquires its input data are independent of the means by which another acquires its data. When used with a(n): A Combining sets of data with sets of instructions B An example of "multiple data, multiple instruction" C D is for "data alone," meaning there is just one instruction. One order, several results