Q4. pts] Program Analysis - Please analyze the value of following registers after executing the instruction in the same line. # Calculate the value of myEg = 1+x+ x2+ x3 (x =5) .data word var x: . Word 3 var n: word 0 myEg: .text 19lobl main: main $s0,var x Sslavar n addi $s2, $zero, 1 addi $t0,$zero, 1 # $s0 = lw # $s1 # $s2 = # $t0 = $s2s2,$s0 Sto Sto, $s2 # $s2 = # St0 = mul add $s2 = # $t0 = Ss2Ss2, $s0 Sto, $to,$s2 mul add # $s2 $s2s2, $s0 $t0, $to, $s2 mul $t0 add $t0myEa $v0,10 SW li main
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- Convert 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.translate the following MIPS code to C. Assume that the variables i, j, and k areassigned to the registers $s0, $s1, and $s2, respectively. Assume that the baseaddress of the array A is in registers $s6.Loop: blt $s0, $s1, Exitbge $s1, $s2, Exitaddi $s1, $s1, 5j LoopExit:addi $t0, $zero, 4ble $s0, $t0, Donesll $t1, $s0, 2add $t2, $s6, $t1sw $zero, 0($t2)Done:Modify the following program in lc-3 to get the input(numbers) from the keyboard and print the result to console. .ORIG x3000Loop LD R0, number1 ; load number1 into R0LDR R1, number2 ; load number2 into R1ST R2, SaveR2 ; save register R2LD R5, goSUB ; load address of SUB into R4JSRR R5 ; go to subroutine whose address in R5STR R3, result ; store resultLD R2, SaveR2 ; restore old value R2HALTnumber1 .FILL #10number2 .FILL # -8goSUB .FILL SUB ; initialize goSUB to address of SUBSaveR2 .BLKW #1; reserve space SaveR2 and SaveR3result .BLKW #1SUB NOT R1, R1ADD R2, R1, #1ADD R3, R0, R2RET.END
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