Question: The question to the problem. n pow.s you will implement a main function and a pow function. pow will take two integers ( let s

The question to the problem.
n pow.s you will implement a main function and a pow function. pow will take two
integers (lets call them n and m) and return nm . You must compute nm using recursion.
Can you find a smaller power computation inside 25? What is the base case? What is
the recursive case?
.global main
.type main, %function
main:
sub sp, sp, #16
str lr,[sp]
//we'll store two ints at [sp, #8] and [sp, #12]
ldr x0,=int_scanf
add x1, sp, #8
add x2, sp, #12
bl scanf
//===============
// don't touch anything above this line.
// Fix: call pow using the two ints on the stack as parameters the first parameter is at sp+8 and the second is at sp+12
//Fix: Print out the results using the int_printf format
mov w0, #0
ldr lr,[sp]
add sp, sp, #16
ret
.global pow
.type pow, %function
pow:
//fix: implement the recursive function `int pow(int n, int m)` that returns n^m.
.data
int_scanf: .asciz "%d %d"
int_printf: .asciz "%d
"
Here is my code that I had already implemented.
.global main
.type main, %function
main:
sub sp, sp, #16// Allocate space on stack
str lr,[sp]// Save return address
// Read two integers (n and m) from input
ldr x0,=int_scanf
add x1, sp, #8// Address for 'n'
add x2, sp, #12// Address for 'm'
bl scanf
// Load n and m from the stack and call pow(n, m)
ldr w0,[sp, #8]// Load n
ldr w1,[sp, #12]// Load m
bl pow // Call pow(n, m), result in w0
// Print the result using printf
ldr x0,=int_printf
mov w1, w0// Move result to w1 for printing
bl printf
// Clean up and return
ldr lr,[sp]// Restore return address
add sp, sp, #16// Deallocate stack space
ret
.global pow
.type pow, %function
pow:
sub sp, sp, #16// Allocate stack space for recursion
str lr,[sp, #8]// Save link register
// Base case: if m ==0, return 1
cmp w1, #0
beq base_case
// Recursive case: compute n^(m-1) and multiply by n
sub w1, w1, #1// Decrease m by 1
mov w2, w0// Store n in w2
bl pow // Recursive call pow(n, m-1)
mul w0, w0, w2// Multiply result by n (w2)
// Restore link register and return
ldr lr,[sp, #8]
add sp, sp, #16
ret
base_case:
mov w0, #1// Base case: return 1
ldr lr,[sp, #8]// Restore link register
add sp, sp, #16// Deallocate stack space
ret
.data
int_scanf: .asciz "%d %d"// Format for scanf
int_printf: .asciz "+%d
"// Format for printf

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