The shared variables, x, a, and b, as used in the following command sequence: place a in a register, place b in a register, place x in a register to store the result of the addition of a and b, is a good example of an atomic instruction or transaction.
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- Convert the given code fragment to assembly code fragment, using only instructions of the following types. These instructions are generally discussed in class. Here X,Y,Z are any memory locations; R, R1, R2 are any general registers; L is a label in the code (you can use any names as labels, ex. L, L1, L2 etc. ). load X, R //copy contents of memory location X into R. store R, X //Store contents of R into Mem location X cmp R1, R2 //Compute R1-R2 and update condition codes; //throw away result of subtraction. jmp L //Jump to location L in the code. jmpp L //If P bit is 1, Jump to location L in the code add X, R //Add contents of X,R and store result in R; //Also update the condition codes. Be careful about what type of argument is allowed in the instruction (Memory or Register). Ex. the first argument of ADD instruction is memory, not register. Do Not…Generate a listing file for AddTwoSum.asm(in 3.4.3 of the text) and write a description of the machine code bytes generated for each instruction. You can write your descriptions directly over the .lst file. You might have to guess at some of the meanings of the byte values. ; AddTwoSum.asm - Chapter 3 example. .386.model flat,stdcall.stack 4096ExitProcess proto,dwExitCode:dword .datasum dword 0 .codemain proc mov eax,5 add eax,6 mov sum,eax invoke ExitProcess,0main endpend mainIs an asynchronous or synchronous interface preferable for connecting the CPU to the memory? Justification must be provided.
- In c++ write an assembler in which it will read a program written in HACK assembly language from an external file and ultimatley translate each line of code into the binary equivalent that can be run on the computer I built so based off the following hdl files Computer below others are in the images. CHIP Computer { IN reset; PARTS: //Read-only memory (ROM) for instruction fetch ROM32K(address=PC,out=instruction); //Central Processing Unit (CPU) for instruction executionCPU(instruction=instruction,reset=reset,inM=outMemo,outM=CPUoutM,writeM=wM,addressM=adM,pc=PC); //Memory for data storage and control logic Memory(in=CPUoutM,load=wM,address=adM,out=outMemo); }int i = 5; is a statement in a C program. A. during execution, value of i may change but not its address B. during execution both the address and value may change C. repeated execution may result in different addresses for i D. i may not have an associated addressQuiz 5: In this problem we want to set the control signals of the datapath shown below (also in in slide # 1 of "chapter3_single_cycle_datapaths.pptx") so that it supports execution of a new instruction called swi. Single Cycle Datapath: PC Read Instru- address ction [31-0] Instruction memory Sns Add Ins 1 [25-21] 1 [20-16] [15-11]. 1[10-0] RegWrite Read register 1 Read register 2 Write register Write data Read data 1 Read data 2 Read Ins Write 3ns Sign extend 2ns MemWrite Read Read address data Write address Read Gns Write data Write 10ns ins ALUSTO1 MemRead ALU Result 2ns ALUOP1 -XEWO) ins ALUSrc2 ALUSrc3 x=3 ins ALU Result 2ns ALUOP2 swi rd, rs, rt, imm # Memory [R[rs]]= R[rt], R[rd] =R [rs]+R [rt]+Imm #this instruction copies contents of "rt" register into the main memory addressed by the "rs" register. In the same cycle it add "rs" and "rt" register contents along with the "imm" field of the instruction and writes the final result into the "rd" register. You are NOT allowed to…
- When translating conditionals or loops, the generated low-level code contains jumps. For the CPU to be able to execute a jump, the target of a jump must be a valid memory address. However, code generators typically generate jumps to symbolic addresses (labels). For example: beq exit Answer the following questions about symbolic addresses. i. What are the advantages of using symbolic addresses? ii. Which programs are responsible for translating such symbolic addresses to actual memory addresses? iii. How do code generators create symbolic addresses?Suppose we have one floating-point multiplication unit in our microprocessor, and it takes seven cycles for it to complete the multiplication. If we get 3 back-to-back floating-point multiply instructions with no data dependency on the previous values, then what type of Hazard can we run into? a. Exception or Interrups b. Control Hazard c. No Hazard, as there is no dependency on the data between the three instructions e. Data Hazard f. Structural HazardIn the context of semaphores, an atomic operation: options: a must be implemented on a device that has been "hardened" against disruption by cosmic radiation b can be combined with other atomic operations into what are commonly known as molecules c will cause a signal to be thrown if run on a machine with an insufficient number of processors d is a sequence of two or more operations that can only be executed as a single unit
- Code a descriptor that describes a data memory segment that grows upward and begins at location 03000000H and ends at location 05FFFFFFH and can be written. The memory has not been accessed, and can be accessed with the lowest privilege level. Assume that the segment is available (i.e., AV=1) and that the instructions are 32 bits (i.e., D=1).When it says "memory leaks," what exactly are we talking about when it comes to dynamic memory allocation?Computer organization and assembly language Please help me with this. I have to write line by line what each line of codes does. CODE IS BELOW: .model small .386 .stack 100h .data msg1 db 13, 10, "Enter any number --> ", "$" msg2 db "Enter an operation +,- * or / --> ",13, 10, "$" msg3 db "The Operation is --> ", "$" msg4 db "The result is --> ", "$" By_base dd 21 by_10 dd 10 ; 32 bits variable with initial value = 10 sp_counter db 0 ; 8 bits variable with initial value of zero disp_number dd 0 ; 32 bits variable with initial value = 0 disp_number2 dd 0 disp_number3 dd 0 op_type db 0 last_key dd 0 ; 32 bits variable with initial value of zero remainder db 0 .code main proc mov ax,@data;set up datasegment movds,ax mov dx,offset msg1 call display_message callm_keyin calloperation mov dx,offset msg1 calldisplay_message callm_keyin cmpop_type, "+" jnz short skip_plus callop_plus skiP_plus: cmp op_type, "-" jnz short skip_minus callop_minus…