在 x86 Linux 中,如何实现 sin(x)
在使用泰勒展开的汇编代码中?
最佳答案
您没有说明哪种 CPU 架构,所以我假设为 x86。
最简单(也可能是最低效)的方法是在 RPN 中编写公式,它几乎可以直接映射到 FPU 指令。
例子,
代数公式:x - (x^3/3!) + (x^5/5!)
RPN:x x x * x * 3 2 */- x x * x * x * x * 5 4 * 3 * 2 */+
变成:
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
有一些明显的优化策略——
每个术语的 xxxxx 等,存储一个
“运行产品”,然后乘以
每次 x*x
计算每个的阶乘
术语,做同样的“运行产品”
这是 x86 FPU 的一些注释代码,每条 FPU 指令之后的注释显示了该指令执行后的堆栈状态,堆栈顶部 (st0) 在左侧,例如:
fldz ; 0
fld1 ; 1, 0
--剪断--
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
运行程序:
$ ./taylor-sine 0.5 1
0.4791666667
$ ./taylor-sine 0.5 5
0.4794255386
$ echo "s(0.5)"|bc -l
.47942553860420300027
关于使用泰勒展开的 sin(x) 汇编代码,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1252929/