user: 这段汇编代码用C语言表示是什么样的？

```asm
<rffti1>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x78,%rsp
  mov    %edi,-0x64(%rbp)
  mov    %rsi,-0x70(%rbp)
  mov    %rdx,-0x78(%rbp)
  mov    -0x78(%rbp),%rdx
  mov    -0x64(%rbp),%eax
  mov    %rdx,%rsi
  mov    %eax,%edi
  call   <factorize@plt>
  mov    -0x78(%rbp),%rax
  mov    0x4(%rax),%eax
  mov    %eax,-0x34(%rbp)
  movss  D7(%rip),%xmm0
  pxor   %xmm1,%xmm1
  cvtsi2ssl -0x64(%rbp),%xmm1
  divss  %xmm1,%xmm0
  movss  %xmm0,-0x30(%rbp)
  movl   $0x0,-0x38(%rbp)
  mov    -0x34(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x2c(%rbp)
  movl   $0x1,-0x44(%rbp)
  cmpl   $0x0,-0x2c(%rbp)
  je     L51
  movl   $0x1,-0x48(%rbp)
  jmp    L56
L52:
  mov    -0x48(%rbp),%eax
  cltq
  add    $0x1,%rax
  lea    0x0(,%rax,4),%rdx
  mov    -0x78(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%eax
  mov    %eax,-0x28(%rbp)
  movl   $0x0,-0x40(%rbp)
  mov    -0x44(%rbp),%eax
  imul   -0x28(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x64(%rbp),%eax
  cltd
  idivl  -0x24(%rbp)
  mov    %eax,-0x20(%rbp)
  mov    -0x28(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x1c(%rbp)
  movl   $0x1,-0x4c(%rbp)
  jmp    L32
L64:
  mov    -0x44(%rbp),%eax
  add    %eax,-0x40(%rbp)
  mov    -0x38(%rbp),%eax
  mov    %eax,-0x50(%rbp)
  pxor   %xmm0,%xmm0
  cvtsi2ssl -0x40(%rbp),%xmm0
  movss  -0x30(%rbp),%xmm1
  mulss  %xmm1,%xmm0
  movss  %xmm0,-0x18(%rbp)
  pxor   %xmm0,%xmm0
  movss  %xmm0,-0x54(%rbp)
  movl   $0x3,-0x3c(%rbp)
  jmp    L60
L49:
  addl   $0x2,-0x50(%rbp)
  movss  -0x54(%rbp),%xmm1
  movss  D14(%rip),%xmm0
  addss  %xmm1,%xmm0
  movss  %xmm0,-0x54(%rbp)
  movss  -0x54(%rbp),%xmm0
  mulss  -0x18(%rbp),%xmm0
  movss  %xmm0,-0x14(%rbp)
  mov    -0x50(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x8(%rax),%rdx
  mov    -0x70(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  call   <cos@plt>
  movd   %xmm0,%eax
  mov    %eax,(%rbx)
  mov    -0x50(%rbp),%eax
  cltq
  shl    $0x2,%rax
  lea    -0x4(%rax),%rdx
  mov    -0x70(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x14(%rbp),%eax
  movd   %eax,%xmm0
  call   <sin@plt>
  movd   %xmm0,%eax
  mov    %eax,(%rbx)
  addl   $0x2,-0x3c(%rbp)
L60:
  mov    -0x3c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L49
  mov    -0x20(%rbp),%eax
  add    %eax,-0x38(%rbp)
  addl   $0x1,-0x4c(%rbp)
L32:
  mov    -0x4c(%rbp),%eax
  cmp    -0x1c(%rbp),%eax
  jle    L64
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x44(%rbp)
  addl   $0x1,-0x48(%rbp)
L56:
  mov    -0x48(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jle    L52
  jmp    L62
L51:
  nop
L62:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D7
type_infered: f32
name: retrieve_data
arguments: tag: D14
type_infered: byte[4]

tool: D7: 6.28318530717959

tool: {"D14": [0x00,0x00,0x80,0x3f]}

assistant: ```c
void rffti1(int n, float wa[], int ifac[13 + 2]) {
  static const float twopi = 6.28318530717959;
  float arg, argh, argld, fi;
  int i, j;
  int k1, l1, l2;
  int ld, ii, nf, ip, is;
  int ido, ipm, nfm1;
  factorize(n, ifac);
  nf = ifac[1];
  argh = twopi / n;
  is = 0;
  nfm1 = nf - 1;
  l1 = 1;
  if (nfm1 == 0)
    return;
  for (k1 = 1; k1 <= nfm1; k1++) {
    ip = ifac[k1 + 1];
    ld = 0;
    l2 = l1 * ip;
    ido = n / l2;
    ipm = ip - 1;
    for (j = 1; j <= ipm; ++j) {
      ld += l1;
      i = is;
      argld = (float)ld * argh;
      fi = 0;
      for (ii = 3; ii <= ido; ii += 2) {
        i += 2;
        fi += 1;
        arg = fi * argld;
        wa[i - 2] = cos(arg);
        wa[i - 1] = sin(arg);
      }
      is += ido;
    }
    l1 = l2;
  }
}
```