In x86 Linux, how can I implement sin(x)
in assembly code using Taylor Expansion?
问题:
回答1:
Would this article help you?
http://www.coranac.com/2009/07/sines/
It has a couple of algorithms for computing approximate sin(x) values, with both C and assembly versions. Granted, it's ARM assembly, but the gist of it should translate easily to x86 or similar.
回答2:
You don't state which CPU architecture so I'm assuming x86.
The simplist (and possibly most inefficient) way would be to write the formula in RPN, which can be mapped almost directly to FPU instructions.
Example,
algebraic formula : x - (x^3/3!) + (x^5/5!)
RPN : x x x * x * 3 2 * / - x x * x * x * x * 5 4 * 3 * 2 * / +
which becomes :
fld x
fld x
fld x
fmul
fld x
fmul
fild [const_3]
fild [const_2]
fmul
fdiv
fsub
fld x
fld x
fmul
fld x
fmul
fld x
fmul
fld x
fmul
fild [const_5]
fild [const_4]
fmul
fild [const_3]
fmul
fild [const_2]
fmul
fdiv
fadd
There are some obvious optimisation strategies -
- instead of calculating x, xxx, xxxxx etc for each term, store a 'running product' and just multiply by x*x each time
- instead of calculating the factorial for each term, do the same 'running product'
Here's some commented code for x86 FPU, the comments after each FPU instruction show the stack state after that instruction has executed, with the stack top (st0) on the left, eg :
fldz ; 0
fld1 ; 1, 0
--snip--
bits 32
section .text
extern printf
extern atof
extern atoi
extern puts
global main
taylor_sin:
push eax
push ecx
; input :
; st(0) = x, value to approximate sin(x) of
; [esp+12] = number of taylor series terms
; variables we'll use :
; s = sum of all terms (final result)
; x = value we want to take the sin of
; fi = factorial index (1, 3, 5, 7, ...)
; fc = factorial current (1, 6, 120, 5040, ...)
; n = numerator of term (x, x^3, x^5, x^7, ...)
; setup state for each iteration (term)
fldz ; s x
fxch st1 ; x s
fld1 ; fi x s
fld1 ; fc fi x s
fld st2 ; n fc fi x s
; first term
fld st1 ; fc n fc fi x s
fdivr st0,st1 ; r n fc fi x s
faddp st5,st0 ; n fc fi x s
; loop through each term
mov ecx,[esp+12] ; number of terms
xor eax,eax ; zero add/sub counter
loop_term:
; calculate next odd factorial
fld1 ; 1 n fc fi x s
faddp st3 ; n fc fi x s
fld st2 ; fi n fc fi x s
fmulp st2,st0
fld1 ; 1 n fc fi x s
faddp st3 ; n fc fi x s
fld st2 ; fi n fc fi x s
fmulp st2,st0 ; n fc fi x s
; calculate next odd power of x
fmul st0,st3 ; n*x fc fi x s
fmul st0,st3 ; n*x*x fc fi x s
; divide power by factorial
fld st1 ; fc n fc fi x s
fdivr st0,st1 ; r n fc fi x s
; check if we need to add or subtract this term
test eax,1
jnz odd_term
fsubp st5,st0 ; n fc fi x s
jmp skip
odd_term:
; accumulate result
faddp st5,st0 ; n fc fi x s
skip:
inc eax ; increment add/sub counter
loop loop_term
; unstack work variables
fstp st0
fstp st0
fstp st0
fstp st0
; result is in st(0)
pop ecx
pop eax
ret
main:
; check if we have 2 command-line args
mov eax, [esp+4]
cmp eax, 3
jnz error
; get arg 1 - value to calc sin of
mov ebx, [esp+8]
push dword [ebx+4]
call atof
add esp, 4
; get arg 2 - number of taylor series terms
mov ebx, [esp+8]
push dword [ebx+8]
call atoi
add esp, 4
; do the taylor series approximation
push eax
call taylor_sin
add esp, 4
; output result
sub esp, 8
fstp qword [esp]
push format
call printf
add esp,12
; return to libc
xor eax,eax
ret
error:
push error_message
call puts
add esp,4
mov eax,1
ret
section .data
error_message: db "syntax: <x> <terms>",0
format: db "%0.10f",10,0
running the program :
$ ./taylor-sine 0.5 1
0.4791666667
$ ./taylor-sine 0.5 5
0.4794255386
$ echo "s(0.5)"|bc -l
.47942553860420300027
回答3:
Why? There is already a FCOS and FSIN opcode since the 80387 (circa 1987) processor
source:
http://ref.x86asm.net/coder32.html
Wikipedia
Personal friend from demo scene
回答4:
long double LD_COS(long double __X)
{
register long double __VALUE;
__asm__ __volatile__(
"fcos \n\t"
: "=t" (__VALUE)
: "0" (__X)
);
return __VALUE;
}
long double LD_SIN(long double __X)
{
register long double __VALUE;
__asm__ __volatile__(
"fsin \n\t"
: "=t" (__VALUE)
: "0" (__X)
);
return __VALUE;
}