Assuming that variables i and j are stored in registers $s0 and $s1 and integer array A is located at the address stored in register $s2, translate the following code snippets to MIPS assembly language: 1. if(i == j) A[0] = i; else 2. A[0] =j; for(i=0; i
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- Using the information below, in assembly language: x BYTE -2, -3, 2, 1 z WORD 3000h, 4000h, 5000h, 6000h y WORD -14, 32 bx starts with a value of 2222h, what is the hex value of bx when the following instructions execute in sequence? 1. mov bl, x 2. mov bh, [x+4]Write an assembly code to implement the y = (x1+x2) * (x3+x4) expression on 2-address machine, and then display the value of y on the screen. Assume that the values of the variables are known. Hence, do not worry about their values in your code. The assembly instructions that are available in this machine are the following: Load b, a Load the value of a to b Add b, a Add the value of a to the value of b and place the result in b Subt b, a Subtract the value of a from the value of b and place the result in b Mult b, a Multiply the values found in a and b and place the result in b Store b, a Store the value of a in b. Output a Display the value of a on the screen Halt Stop the program Note that a or b could be either a register or a variable. Moreover, you can use the temporary registers R1 & R2 in your instructions to prevent changing the values of the variables (x1,x2,x3,x4) in the expression. In accordance…Assuming the following MIPS assembly language code snippet is loaded into the memory of a MIPS-32 processor-based system starting at memory address 0x00400000. Fill the blanks on the right with the corresponding machine code in hexadecimal. Loop: sll $t1, $s3, 2 0x[a] add $t1, $t1, $s6 0x[b] lw $t0, 0($t1) 0x[c] bne $t0, $s5, Exit 0x[d] addi $s3, $s3, 1 0x[e] j Loop 0x[f] Exit: Note: Do NOT include '0x' and do NOT omit the leading '0's in your answers. Please show work by hand not just by running MARS and showing answers there please!!
- Write and simulate a MIPS assembly-language routine that: 1. Prints your group number, for example “Group 1”, 2. Computes the dot product of two vectors, A_vec and B_vec, as described in Lab 4 of the Lab Manual, 3. Stores the result at memory word DOTPROD, and 4. Prints the result preceded by the phrase: “The result of the dot product is: “. Your data segment should look like the following: .data group: .asciiz “\nGroup x” msg: .asciiz "\nThe result of the dot product is: " A_vec: .word x, 4, 20, 13, 3, 10, 5 B_vec: .word 4, 2, 1, 2, 5, 2, 4 n: .word 7 DOTPROD: .word 0 Replace each “x” with your group number.1. T/F - if (B)=006000 (PC)=003600 (X)=000090, for the machine instruction 0x032026, the target address is 003000.2. T/F – PC register stores the return address for subroutine jump.3. T/F – S register contains a variety of information such as condition code.4. T/F – INPUT WORD 1034 – This means Operating system should reserve 1034 bytes in memory5. T/F - In a two pass assembler, adding literals to literal table and address resolution of local symbol are done using first pass and second pass respectively.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…
- Consider the following MIPS assembly language program: # Data area.datamydata: .word 0x4CAFE, 0xBABE#output: .word 0 # for storing the result## Instructions (program) area.textmain: la $s0, mydata # load the address of memory location mydata to $s0lw $t0, ($s0) #lw $t1, 4($s0) #sub $t2, $t0, $t1 ##la $s1, output # Load the address of memory location output to $s1sw $t2, ($s1) ##li $v0, 1 # syscall #1 (print int)srl $a0, $t2, 1 # divide by 2syscall # invoke syscall to print # # The END 1) What will be seen on MARS console when this program runs? [x]3. Create an 80x86 ASSEMBLY LANGUAGE program that define an array of doubleword numbers then read twovalues, first one indicates how many actual numbers will be in the array, and an integer n. Theprogram should display all of the numbers in the array that are greater than the number n. MUST PROVIDE FULL CODE AND SCREENSHOT OF OUTPUT PLEASEConsider memory storage of a 32-bit word stored at memory word 34 in a byte addressable memory. (a) What is the byte address of memory word 34? (b) What are the byte addresses that memory word 34 spans? (c) Draw the number 0x3F526372 stored at word 342 in both big-endian and little-endian machines. Clearly label the byte address corresponding to each data byte value.
- Write a Java program that prompts the user for the page size used in a virtual memory system; this will be a power of two between 512 (29) and 16384 (214), inclusive. Your program should check the user input for page size to make sure it is one of the allowable inputs (must be a power of 2 and cannot be smaller than 512 or larger than 16384), and should then prompt the user for a virtual address (assume 32-bit virtual addressing, so your code must be able to accept any input between 0 and 4294967295, which is 232-1). Given this input, the program must output the virtual page number and the offset within the page. Sample output might look like this: Please enter the system page size: 1024 Please enter the virtual address: 10000 This address is in virtual page: 9 At offset: 784Convert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.Suppose you have 100B memory space where data is stored and you wish to search a data value of 0AH within the memory space. Write an assemblylanguage code to do the task in EMU 8086 microprocessor