Question

Write an assembly language program to calculate the following.

(((((2⁰ + 2¹)X 2²)+ 2³)X 2⁴)+ 2⁵...)+ 2ⁿ

Hint:Note that when n is even, carrying result is multiplied by 2 n . When n is odd, carrying result is added to 2 n (above formula shows the case where n is odd).

As an example, when n= 5, it should print 352.

•Your test caller should do the followings.

– Read n from keyboard.

– Call the sub procedure to do the calculation.

– Print the appropriate value, if possible or display appropriate messages for returned values -1 and -2 (see below)

•Your callee should do the followings

–Procedure should take n as a parameter.

–If n ≥ 0 , procedure should calculate the correct value. if n <0, -1 should be returned.

–As n increases, at one point, you will not be able to hold the result in a 32-bit register. You should check for it (probably using jo instruction). You should return -2 in that case.

Answer 1

Please Refer Comments for help.

f: /*Called Function */

push rbp

mov rbp, rsp

mov DWORD PTR [rbp-20], edi

cmp DWORD PTR [rbp-20], 0 /*comparing n with 0 */

jns .L2

mov eax, -1

jmp .L3

.L2:

mov QWORD PTR [rbp-8], 20 /*initializing result with 20 */

mov DWORD PTR [rbp-12], 21 /* starting with 21 */

jmp .L4

.L5:

mov eax, DWORD PTR [rbp-12]

cdqe

add QWORD PTR [rbp-8], rax

add DWORD PTR [rbp-12], 1

mov eax, DWORD PTR [rbp-12]

cdqe

mov rdx, QWORD PTR [rbp-8]

imul rax, rdx /*multily with even no. */

mov QWORD PTR [rbp-8], rax

add DWORD PTR [rbp-12], 1 /* increment iterator */

mov eax, 2147483648 /* Check for range of 32 bit integer */

cmp QWORD PTR [rbp-8], rax

jl .L4

mov eax, -2

jmp .L3

.L4:

mov eax, DWORD PTR [rbp-20]

add eax, eax

cmp DWORD PTR [rbp-12], eax

jle .L5

mov rax, QWORD PTR [rbp-8]

.L3:

pop rbp

ret

.LC0:

.string "%d"

main: /* Main Method */

push rbp

mov rbp, rsp

sub rsp, 16

lea rax, [rbp-4]

mov rsi, rax

mov edi, OFFSET FLAT:.LC0

mov eax, 0

call __isoc99_scanf

mov eax, DWORD PTR [rbp-4]

mov edi, eax

call f

mov esi, eax

mov edi, OFFSET FLAT:.LC0

mov eax, 0

call printf /* Print Result */

nop

leave

ret

Please give feedback for my answer if it helped you or raise your query in case of any confusion.

Happy coding...