Register Indirect Addressing C.f. indirect addressing EA = (R) %3D Operand is in memory cell pointed to by contents of register R Large address space (2n) One fewer memory access than indirect addressing DRAW THE DIAGRAM???

Register Indirect Addressing C.f. indirect addressing EA = (R) %3D Operand is in memory cell pointed to by contents of register R Large address space (2n) One fewer memory access than indirect addressing DRAW THE DIAGRAM???

Name: Register Indirect Addressing• .f. indirect addressingEA = (R)Operand
Uploaded: 2024-03-20T06:51:23.000Z
Channel: Bartleby
Description: Register Indirect Addressing• .f. indirect addressingEA = (R)Operand is in memory cell pointed to bycontents of register RLarge address space (2n)One fewer memory access than indirectaddressing• DRAW THE DIAGRAM???

Systems Architecture

7th Edition

ISBN:9781305080195

Author:Stephen D. Burd

Publisher:Stephen D. Burd

Chapter11: Operating Systems

Section: Chapter Questions

Problem 30VE

See similar textbooks

Similar questions

B) Amain memory address is viewed as consisting of multiple fields. List and define these fields for Direct Mapped Cache, Associative Mapped Cache and Set-associative Mapped Cache Memory.
6. Assume that two numbers: dividend and divisor are saved in memory address M1 and M2 respectively. Quotient and remainder should be saved in R1 and R2 respectively. Write assembly language instructions and then list microoperations for each instruction and list the control signals required to be activated for each microoperation. MBR is used as buffer for any register to register transfer operation. Signal Description: Control signals operation Comments C0 MAR to RAM (through address bus) C1 PC to MBR C2 PC to MAR C3 MBR to PC C4 MBR to IR C5 RAM to MBR C6 MBR to ALU C7 Accumulator to ALU C8 IR to MAR C9 ALU to Accumulator C10 MBR to Accumulator C11 Accumulator to MBR C12 MBR to RAM (through data bus) C13 IR to Control Unit C14 MBR to R1 C15 MBR to R2 C16 MBR to R3 C17 MBR to R4…
Fill in blank Suppose that segmentation technique is used for memory management. If the memory addresses are 7-bit binary numbers where the first two bits are used to specify segment ID, then the size of a user stack will be ---------- bytes.
Please solve, Topic: Microprocessor Q1a) AL= 73 CL=29 ADD AL, CL DAA What is the value of CF after execution? 0 1 Q 1b) MOV CX, [481d] ; assuming DS= 2162H, logical address will be?
1. Name all of the general purpose registers and some of their special functions. 2. How are the segment registers used to form a 20-bit address? 3. (a) If CS contains 03E0H and IP contains 1F20H, from what address is the next instruction fetched? (b) If SS contains 0400H and SP contains 3FFEH, where is the top of the stack located? (c) If a data segment deigns at address 24000H, what is the address of the last location in the segment? 4. Explain what the instruction array and data caches are used for. 5. What is the EU and BIU, and what purpose in the microcomputer? 6. Two memory locations, beginning at address 3000H, contain the bytes 34H and 12H. What is the word stored at location 3000H? See Figure 2.26 for details. Address 3000 Data 34 3001 12 Figure 2.26 For question 6 7. What is a physical address? What are the differences between the 8086 logical and physical memory maps? 8. May memory segments overlap? If so, what is the minimum number of overlapped bytes…
SUBJECT: COMPUTER ORGANASATION 1. a. How many address bits are required to access 256K words of memory? b. How many bytes of memory can be addressed by 24 address bits?
Memory address translation is useful only if the total size of virtual memory (summed over all processes) needs to be larger than physical memory. True or False. Justify your answer.
1. The hypothetical machine has two I/O instructions: 0011 = Load AC from I/O 0111 = Store AC to I/O In these cases, the 12-bit address identifies a particular I/O device. List the steps for every execution for the following program and illustrate using table that explain the process below : a. Load AC from device 5. b. Add contents of memory location 940. c. Store AC to device 6. d. Assume that the next value retrieved from device 5 is 3 and that location 940 contains a value of 2. Please pointing a, b,c ans. Because one I already upload this question and I didn't understand which one is and of a...please write ans a, b , c please
witch of the following is the mode of addressing specifies the index of a register where the value of interest is located: A immediate addressing B pseudo-direct addressing C base (or displacement) addressing D register addressing E PC-relative addressing
1) For a Pentium II descriptor that contains a base address of 0004B100H, a limit of 00FFFH, and G = 1, what starting and ending locations are addressed by this descriptor? 2) Code a descriptor that describes a memory segment that begins at location 0005CF00h and ends at location 00060EFFh. The memory segment is a data segment that grows upward in the memory system and can be written. The segment has a user level privilege (lowest) and has not been accessed. The descriptor is for an 80386 microprocessor.
Using indirect addressing mode (like for LDI and STI), the first memory location contains the __________________ of the memory location to load data from or store data to.
The following diagram shows some registers like processor registers R1 and R2, Program counter PC and Index Register XR along with their corresponding values. It also shows a memory with some instructions like instruction A and next instruction. The memory holds instruction B which consists of four fields as given above. First field of instruction represents the addressing Mode (I), second field specifies Opcode (operation code) ADD representing operation addition, the third field represents Address field 1 and the fourth field represents Address field 2. Consider the following addressing modes, evaluate the result of execution of above instruction by giving steps of evaluation for each addressing mode for the scenario given above. Ø Immediate Mode Ø Direct Mode Ø Register Ø Relative Mode Ø Index Mode Choose your own values for variables (v – w), T1, T2. Choose any one of the given value for T3 (200 or 300). V=700 W=800 T1=200 T2=200 T3=300