Using SSE to mimic the standard Math.pow function - c

Question

I'm trying to learn how to work with SSE and I decided to realize a simple code that computes n^d, using a function that gets called by a C program.
Here's my NASM code:
section .data
resmsg: db '%d^%d = %d', 0
section .bss
section .text
extern printf
; ------------------------------------------------------------
; Function called from a c program, I only use n and d parameters but I left the others
; ------------------------------------------------------------
global main
T equ 8
n equ 12
d equ 16
m equ 20
Sid equ 24
Sn equ 28
main:
; ------------------------------------------------------------
; Function enter sequence
; ------------------------------------------------------------
push ebp ; save Base Pointer
mov ebp, esp ; Move Base Point to current frame
sub esp, 8 ; reserve space for two local vars
push ebx ; save some registries (don't know if needed)
push esi
push edi
; ------------------------------------------------------------
; copy function's parameters to registries from stack
; ------------------------------------------------------------
mov eax, [ebp+T] ; T
mov ebx, [ebp+n] ; n
mov ecx, [ebp+d] ; d
mov edx, [ebp+m] ; m
mov esi, [ebp+Sid] ; Sid
mov edi, [ebp+Sn] ; Sn
mov [ebp-8], ecx ; copy ecx into one of the local vars
;
; pow is computed by doing n*n d times
;
movss xmm0, [ebp+n] ; base
movss xmm1, [ebp+n] ; another copy of the base because xmm0 will be overwritten by the result
loop: mulss xmm0, xmm1 ; scalar mult from sse
dec ecx ; counter--
cmp ecx,0 ; check if counter is 0 to end loop
jnz loop ;
;
; let's store the result in eax by moving it to the stack and then copying to the registry (we use the other local var as support)
;
movss [ebp-4], xmm0
mov eax, [ebp-4]
;
; Print using C's printf
;
push eax ; result
mov ecx, [ebp-8] ; copy the original d back since we used it as loop's counter
push ecx ; exponent
push ebx ; base
push resmsg ; string format
call printf ; printf call
add esp, 24 ; clean the stack from both our local and printf's vars
; ------------------------------------------------------------
; Function exit sequence
; ------------------------------------------------------------
pop edi ; restore the registries
pop esi
pop ebx
mov esp, ebp ; restore the Stack Pointer
pop ebp ; restore the Base Pointer
ret ; get back to C program
Now, what I'd expect is it to print
4^2 = 16
but, instead, I got
4^2 = 0
I've spent my whole afternoon on this and I couldn't find a solution, do you have any hints?
EDIT:
Since it seems a format problem, I tried converting the data using
movss [ebp-4], xmm0
fld dword [ebp-4]
mov eax, dword [ebp-4]
instead of
movss [ebp-4], xmm0
mov eax, [ebp-4]
but I got the same result.

Answer 1

MOVSS moves single precision floats (32-bit). I assume that n is an integer so you can't load it into a XMM register with MOVSS. Use CVTSI2SS instead. printf cannot process single precision floats, which would converted to doubles by the compiler. It's convenient to use CVTSS2SI at this point. So the code should look like:
...
;
; pow is computed by doing n*n d times
;
cvtsi2ss xmm0, [ebp+n] ; load integer
sub ecx, 1 ; first step (n^1) is done
cvtsi2ss xmm1, [ebp+n] ; load integer
loop:
mulss xmm0, xmm1 ; scalar mult from sse
sub ecx, 1
jnz loop
cvtss2si eax, xmm0 ; result as integer
;
; Print using C's printf
;
push eax ; result
mov ecx, [ebp-8] ; copy the original d back since we used it as loop's counter
push ecx ; exponent
push ebx ; base
push resmsg ; string format
call printf ; printf call
add esp, 16 ; clean the stack only from printf's vars
...