dont use ai to do it !!! Q1). Suppose $s0 stores the base address of word array A and $t0 is associated with m, convert the following instruction into MIPS. A[240] = A[240+m]Q2). Suppose $t0 stores the base address of word array A and $s0 is associated with m, convert the following instruction into MIPS. m= 0 while (m <= 10): A[m] = A[m+4]*6 m = m + 5
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dont use
Q1). Suppose $s0 stores the base address of word array A and $t0 is associated with m, convert the following instruction into MIPS.
A[240] = A[240+m]
Q2).
Suppose $t0 stores the base address of word array A and $s0 is associated with m, convert the following instruction into MIPS.
m= 0
while (m <= 10):
A[m] = A[m+4]*6
m = m + 5
Step by step
Solved in 1 steps
- 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:Which of the following is NOT true of the C program construct a[b] ? Group of answer choices a) The array index should be computed as an rval b) If a is an array of int on MIPS starting at the address 0x800, then a[1] is at 0x804 c) On MIPS, it might use the same unsigned integer add instruction that is used to implement c+d where c and d are both int d) a[b] is not usable as an lval e) On MIPS, one would expect space for a[] to either be allocated on the stack or in the .data segment(Practice) Although the total number of bytes varies from computer to computer, memory sizes of millions and billions of bytes are common. In computer language, the letter M representsthe number 1,048,576, which is 2 raised to the 20th power, and G represents 1,073,741,824, which is 2 raised to the 30th power. Therefore, a memory size of 4 MB is really 4 times 1,048,576 (4,194,304 bytes), and a memory size of 2 GB is really 2 times 1,073,741,824 (2,147,483,648 bytes). Using this information, calculate the actual number of bytes in the following: a. A memory containing 512 MB b. A memory consisting of 512 MB words, where each word consists of 2 bytes c. A memory consisting of 512 MB words, where each word consists of 4 bytes d. A thumb drive that specifies 2 GB e. A disk that specifies 4 GB f. A disk that specifies 8 GB
- I need an assembly program to compute the following expressions x86 masm assembly language it needs to have a DWORD array named ‘z’ of size 3 using DUP operator. Leave the array ‘z’ uninitialized.You can denote the items in the array as [z0 , z1 , z2 ], where z0 is the first item, z1 is the second item, z2 is the third item- Update each array item using the following expressions. z0 = x + y + rz1 = z0 + (y − r)z2 = z0 + (z1 + y) - Where x, y, r are 16-bit integer memory variables. - x = 10, y = 15, r = 4 - usе mоv, mоvzх, mоvsх, аԁԁ, sub іոstruсtіоոs оոly. - dо ոоt аltеr thе vаluе оf х, y аոԁ r ԁurіոg thе соmрutаtіоո. Тrаոsfеr thеm tо аррrорrіаtе rеgіstеrs tо ԁо соmрutаtіоո)- аt thе еոԁ, ореո mеmоry wіոԁоw tо sее thе vаrіаblе z stоrеԁ іո mеmоry (lіttlе еոԁіаո fоrmаt). - usе thе ԁеbuggеr tо vеrіfy yоur аոswеr. sсrееnshот fоr thе соdе аnd rеsulт nееdеdI need an assembly program to compute the following expressions x86 masm assembly language it needs to have a DWORD array named ‘z’ of size 3 using DUP operator. Leave the array ‘z’ uninitialized.You can denote the items in the array as [z0 , z1 , z2 ], where z0 is the first item, z1 is the second item, z2 is the third item- Update each array item using the following expressions. z0 = x + y + rz1 = z0 + (y − r)z2 = z0 + (z1 + y) - Where x, y, r are 16-bit integer memory variables. - x = 10, y = 15, r = 4- Use mov, movzx, movsx, add, sub instructions only. - Do not alter the value of x, y and r during the computation. Transfer them to appropriate registers to do computation)- At the end, open memory window to see the variable z stored in memory (little endian format).- Use the debugger to verify your answer. SCREENSHOT for the CODE AND RESULT NEEDEDRearrange the following code to minimize the total number of cycles, assuming that a dependent instruction following the load will need two clock cycles of delay before getting the data. Load r1, 64 (r2)Add r2, r2, r2Sub r3, r4, r1Load r4, 32 (r4)
- 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 pleaseConvert 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.Solve the 8085 Write a program to load twenty memory locations starting from 8005H, where each location's content should increases by 2 over the previous one, however, the first location should contain 04H, assuming that the programs start at memory location 9009H?
- Pls dont use ai Q1). Suppose $s0 stores the base address of word array A and $t0 is associated with m, convert the following instruction into MIPS. A[240] = A[240+m]a) Fill in the blanks..data A: .word 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 i: .word 7 B: .word 40 45 50 55 60 65 k: .word 1 m: .word 5 # how many f: .word 15 #size of A g: .word 6 #size of B .text .globl main main: # load registers for array A la $s0, A # load address of A lw $s2, i # load word i into s2 # load registers for array B la $s1, B # load address of B lw $s3, k # load word k into s3 #how many to overwrite lw $s4, m # load word g (how many) into s4 # initialize registers to get A[i] sll $s6, $s2, 2 add $s6, $s6, ____ # address of A[i] (new base) # initialize registers to get B[k] sll $s7, $s3, 2 add $s7, $s7, ____ # address of B (new base) #what is the A index to stop? Let's put it into s5 ______________________________ #Loop through A and B Loop: # use slt to compare $s5 and $s2, place the result in t0 ___________________________ beq $t0, $zero, end # load A[i] and store into B[k] lw $t0, 0($s6) sw $t0, 0($s7) #update of the indices addi $s3, $s3, ____ addi $s2,…Describe what each instruction from the following program will actually do: TITLE Task INCLUDE Irvine32.inc .data str1 BYTE "The array sum is: ",0 start BYTE "Enter the Starting Index: ",0 endinx BYTE "Enter the Ending Index: ",0 array DWORD 4, 6, 2, 5, 6, 7, 8, 4 sum DWORD ? j DWORD ? k DWORD ? .code main PROC mov esi, OFFSET array mov ecx, LENGTHOF array mov edx, OFFSET start call WriteString call ReadInt mov j, eax mov esi, j mov edx, OFFSET endinx call WriteString call ReadInt mov k, eax mov ecx, k call ArraySum mov sum,eax call WriteInt main ENDP ArraySum PROC push esi push ecx mov eax, 0 L1: add eax, array[esi] add esi, TYPE DWORD loop L1 pop ecx pop esi ret ArraySum ENDP END main