使用泰勒展开的 sin(x) 汇编代码

Question

在 x86 Linux 中，如何实现 sin(x)在使用泰勒展开的汇编代码中？

Answer 1

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——

而不是计算 x, xxx,
每个术语的 xxxxx 等，存储​​一个
“运行产品”，然后乘以
每次 x*x

代替
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