Question

QT Spim question. Program and answer already given please explaine it.

Please explain the reason why each instruction was used throughout the program step by step given the prompt, what is its purpose to achive the goal asked in the prompt, why is written in that order.

Please not just write to the side what each instruction means in words like load $t0 to $t1. I want to understand the program thoroughly.

Thanks in advance

The preceding program was rather long, because we did not use a loop to make it more compact. Using the same data: numl: .word 34527 num2: .word 98564 num3: .word 12953 num4: .word 68577 Rewrite the program to reverse the order of the numbers and print out as before, but use two short loops to (a) store the words in the stack, and (b) print them out. As in Program 2, you still need to output a CR/LF between numbers, so that each appears on a new line. Hints: You will need to load the address of the first number (numl) into a register, and then address it (and the other numbers) by the indirect- register-plus-offset method. You will need a counter for each loop to determine when you have stored (and later output) four numbers.

Previous information needed to solve problem asked. answe in console is the same for both problems.

Answer 1

Program 2 :-

Explanation :-

On whole if I say, the program is first pushing the elements in order into stack. And, then retrieving it one by one from stack in printing it. So, accordingly, the first element which is pushed into stack is the last one to pop out of the stack. So, if we print the elements one by one by popping out of stack then it will result in printing the elements in reverse order

More detailed Explanation :-
First, we will divide the program into 2 parts :-

1) we will push the elements into stack one by one :-

lw $t0, num1
sub $sp, $sp, 4
sw $t0, 0($sp)

This segment is used to push the word element from memory into stack.
sub $sp, $sp, 4 - is used to create space in stack for 1 word element i.e 4 bytes

So, accordingly we will do the same for all 4 elements.

2) we will pop elements from stack one by one and will print it :-

So, as after push in stack, the order in stack would be :-
bottom->num1->num2->num3->num4->top

So, now the stack pointer will point to num4 first; then num3 then num2 and finally num1

lw $a0,0($sp)
addi $sp,$sp,4
li $v0,1
syscall

This segment is used to pop the word element from stack and print it.
add $sp, $sp, 4 is used to free word space in stack and point to next element.

So, subsequently it is done for each element which will result in printing the elements in reverse order

Program 1 :-

Explanation :-

The process is also same as program 2. Only difference is that it uses loop. In program 2, for are writing the same set of statements 4 times. But, here we take the first memory address of num1 and do the stack operation and then we repeat the same statements for next word address i.e by the adding the memory address with 4. As, it is a word size element. So, each word element takes 4 bytes. So, accordingly, we will add 4 and do the operation in loop