--Must be in assembly language use visual studios console 64 Write a program to
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Question
--Must be in assembly language use visual studios console 64
Write a program to calculate the Nth number in the Fibonacci Sequence. You may use the any of the frameworks provided on blackboard in the frameworks zip. If you use console 32 or console 64 the correct number must be place in the EAX register before you move 0 into EAX to return from the procedure. The N to select which number in the sequence to calculate must be declared in the data segment for console 32 and console 64 applications. Submissions completed by the end of the second week of the assignment will receive 3 bonus points. Submissions completed by the end of the first week of the assignment will receive an additional 3 bonus points. Submissions that use the Windows 32 or Windows 64 to ask the user for N and display the output will receive an addition 4 bonus points.
Explanation / Answer
; fib.asm
;void fib(int32_t *dest, uint32_t count);
; not-unrolled version. See below for a version which avoids all the mov instructions
global fib
fib:
; 64bit SysV register-call ABI:
; args: rdi: output buffer pointer. esi: count (and you can assume the upper32 are zeroed, so using rsi is safe)
;; locals: rsi: endp
;; eax: current edx: prev
;; ecx: tmp
;; all of these are caller-saved in the SysV ABI, like r8-r11
;; so we can use them without push/pop to save/restore them.
;; The Windows ABI is different.
test esi, esi ; test a reg against itself instead of cmp esi, 0
jz .early_out ; count == 0.
mov eax, 1 ; current = 1
xor edx, edx ; prev = 0
lea rsi, [rdi + rsi * 4] ; endp = &out[count]; // loop-end pointer
;; lea is very useful for combining add, shift, and non-destructive operation
;; this is equivalent to shl rsi, 4 / add rsi, rdi
align 16
.loop: ; do {
mov [rdi], eax ; *buf = current
add rdi, 4 ; buf++
lea ecx, [rax + rdx] ; tmp = curr+prev = next_cur
mov edx, eax ; prev = curr
mov eax, ecx ; curr=tmp
;; see below for an unrolled version that doesn't need any reg->reg mov instructions
; you might think this would be faster:
; add edx, eax ; but it isn't
; xchg eax, edx ; This is as slow as 3 mov instructions, but we only needed 2 thanks to using lea
cmp rdi, rsi ; } while(buf < endp);
jb .loop ; jump if (rdi BELOW rsi). unsigned compare
;; the LOOP instruction is very slow, avoid it
.early_out:
ret
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