Read the assembly code below; add comments to explain what each line of code is doing; in one sentence, explain what this procedure is trying to accomplish
new-proc:
sll $a0, $a0, 24
srl $a0, $a0, 24
add $v0, $a0, $zero
jr $ra
new-proc: # new-proc
function
sll $a0, $a0, 24 # shifts a0 left by 24
srl $a0, $a0, 24 # shifts a0 right by 24
add $v0, $a0, $zero # set's $a0 to $v0 i.e,
$v0=$a0
jr $ra # returns
from this function
# this function set's $v0 with first byte of $a0
Read the assembly code below; add comments to explain what each line of code is doing;...
The following MIPS assembly code contains a mistake that violates the MIPS convention in terms of using stack for storing/protecting registers. What is the mistake and how should it be fixed? Correct the corresponding lines in the code. For the corrected code, sketch the stack frame contents at the time when the instruction ‘move $s1, $a1’ is being executed. f: addi $sp, $sp, 12 sw $ra, 8($sp) sw $s1, 4($sp) sw $s0, 0($sp) move $s0, $a0 move $s1, $a1 #...
ASSEMBLY LANGUAGE (Mars MIPS) Starting code: Factorial: #Factorial Recursive function subu $sp, $sp, 4 sw $ra, 4($sp) # save the return address on stack beqz $a0, terminate # test for termination subu $sp, $sp, 4 # do not terminate yet sw $a0, 4($sp) # save the parameter sub $a0, $a0, 1 # will call with a smaller argument jal Factorial # after the termination condition is reached these lines # will be executed lw $t0, 4($sp) # the argument I...
what is the output of the following assembly code ? .data A: .word 84 111 116 97 108 32 105 115 32 . text main:li $v0, 1l li $s0,0 la $s1, A li $t3, 0 loop: slti $t0, $s0, 17 beq: $t0, $zero, end sll $t1, $s0 2 add $t2 $s1 $t1 syscall addi $s0, $s0, 1 j loop end: li $v0, 1 add $a0, $zero, $t3 syscall
What comments would you add to explain the purpose of the code below? How was this code created and when does it get called? private void configureBTN_Click(object sender, EventArgs e) { userMessageLBL.Enabled = userMessageTXT.Enabled; }
. What comments would you add to explain the purpose of the code below? How was this code created and when does it get called? private void fileNameLST_SelectedIndexChanged(object sender, EventArgs e) { attachmentDialog.ShowDialog(); msgTXT.Text = attachmentDialog.SafeFileName; }
WRITE THE FOLLOWING CODE IN FLOATING POINT NUMBERS IN ASSEMBLY LANGUAGE USING MIPS IN MARS .data prompt: .asciiz "\nMaximum number is : " prompt1: .asciiz "\nMinimum number is : " prompt2: .asciiz "\nRange of the array is : " size: .word 10 #load array array: .word 23, -12, 45, -32, 52, -72, 8, 13,22,876 .text #load address of array and size la $s4,array #load address of A lw $t0,size #load i to t0 jal getArrayRange li $v0, 4 la...
Write the assembly code that will transform a squared image into a negative print. (HARD-CODED OUTPUT WILL RESULT IN ZERO.) That is, every pixel whose unsigned value is 0 should change to 255, 1 should change to 254, 2 should change to 253, ... 255 should change to 0. $a0 contains the address of the input buffer, $a1 contains the address of the output buffer address, and $a2 contains the image dimension. Look at the below image for your reference:...
Subroutines in MIPS Determines the minimum of two integers Functions within the MIPS slides describe how one can use subroutines (also called procedures, functions, and methods) in MIPS. Because of the importance of subroutines in modern programming, most hardware designers include mechanisms to help programmers. In a high-level language like C or Java, most of the details of subroutine calling are hidden from the programmer. MIPS has special registers to send information to and from a subroutine. The registers $a0,...
1. Write a program in Assembly language using MIPS instruction set that reads two integer numbers from the user named as start and end number and finds out all the prime numbers between start and end (including start and end). Your program should do the validation of both the numbers as follows: i. start number must be smaller or equal to the end number. ii. Both numbers must be positive. iii. The maximum value for the end number is 10000...