user: 这段汇编代码用C语言表示是什么样的？

```asm
<newtons_method>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  movsd  %xmm0,-0x28(%rbp)
  movsd  %xmm1,-0x30(%rbp)
  movsd  %xmm2,-0x38(%rbp)
  movl   $0x0,-0x1c(%rbp)
  movsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  lea    D50(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  lea    D77(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L16:
  movsd  -0x18(%rbp),%xmm0
  movsd  %xmm0,-0x28(%rbp)
  addl   $0x1,-0x1c(%rbp)
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <f_prime1@plt>
  movq   %xmm0,%rax
  pxor   %xmm0,%xmm0
  movq   %rax,%xmm3
  ucomisd %xmm0,%xmm3
  jp     L26
  pxor   %xmm0,%xmm0
  movq   %rax,%xmm4
  ucomisd %xmm0,%xmm4
  jne    L26
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L69
L26:
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <f1@plt>
  movq   %xmm0,%rbx
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  call   <f_prime1@plt>
  movq   %rbx,%xmm1
  divsd  %xmm0,%xmm1
  movsd  -0x28(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm0
  subsd  -0x28(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <fabs@plt>
  pxor   %xmm0,%xmm0
  ucomisd -0x30(%rbp),%xmm0
  jp     L1
  pxor   %xmm0,%xmm0
  ucomisd -0x30(%rbp),%xmm0
  je     L31
L1:
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x1c(%rbp),%xmm1
  movsd  -0x38(%rbp),%xmm0
  comisd %xmm1,%xmm0
  jb     L31
  mov    $0x1,%edx
  jmp    L64
L31:
  mov    $0x0,%edx
L64:
  cmp    %eax,%edx
  jl     L16
L69:
  lea    D21(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x28(%rbp),%rax
  movq   %rax,%xmm0
  lea    D74(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  movq   %rax,%xmm0
  lea    D115(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x1c(%rbp),%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x30(%rbp),%rax
  movq   %rax,%xmm0
  lea    D106(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x1c(%rbp),%xmm0
  comisd -0x38(%rbp),%xmm0
  jbe    L93
  lea    D60(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  jmp    L9
L93:
  lea    D12(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
L9:
  movsd  -0x18(%rbp),%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: label: D50
data_type: string
name: unpack_data
arguments: label: D0
data_type: string
name: unpack_data
arguments: label: D77
data_type: string
name: unpack_data
arguments: label: D89
data_type: string
name: unpack_data
arguments: label: D21
data_type: string
name: unpack_data
arguments: label: D74
data_type: string
name: unpack_data
arguments: label: D115
data_type: string
name: unpack_data
arguments: label: D52
data_type: string
name: unpack_data
arguments: label: D5
data_type: string
name: unpack_data
arguments: label: D106
data_type: string
name: unpack_data
arguments: label: D60
data_type: string
name: unpack_data
arguments: label: D12
data_type: string

tool: D50: Now calculating 

tool: D0: . 

tool: D77: .. 


tool: D89: f_prime( x_old ) is zero!

tool: D21: Results: 

tool: D74: x_0 = %lf 


tool: D115: x_new = %lf 


tool: D52: Number of iterations = %d 


tool: D5: count = %d 


tool: D106: error_tol = %lf 


tool: D60: Iteration did not convereg.


tool: D12: Iteration did convereg.


assistant: ```c
double newtons_method(double x_0, double error_tol, double max_iters) {
  int count = 0;
  double x_new = x_0;
  printf("Now calculating \n");
  printf(". \n");
  printf(".. \n\n");
  do {
    x_0 = x_new;
    count++;
    if (f_prime1(x_0) == 0) {
      printf("f_prime( x_old ) is zero!\n");
      break;
    }
    x_new = x_0 - f1(x_0) / f_prime1(x_0);
  } while (fabs(x_new - x_0) > (error_tol && (count <= max_iters)));
  printf("Results: \n");
  printf("x_0 = %lf \n", x_0);
  printf("x_new = %lf \n", x_new);
  printf("Number of iterations = %d \n", count);
  printf("count = %d \n", count);
  printf("error_tol = %lf \n", error_tol);
  if (count > max_iters) {
    printf("Iteration did not convereg.\n\n");
  } else {
    printf("Iteration did convereg.\n\n");
  }
  return x_new;
}
```