user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<timename>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %edi,-0x24(%rbp)
  mov    %esi,-0x28(%rbp)
  mov    %edx,-0x2c(%rbp)
  mov    %rcx,-0x38(%rbp)
  pxor   %xmm1,%xmm1
  cvtsi2sdl -0x28(%rbp),%xmm1
  movsd  D11(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  pxor   %xmm0,%xmm0
  cvtsi2sdl -0x2c(%rbp),%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm1
  movsd  D27(%rip),%xmm0
  mulsd  %xmm1,%xmm0
  movsd  D34(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movsd  %xmm0,-0x18(%rbp)
  movsd  -0x18(%rbp),%xmm1
  movsd  D18(%rip),%xmm0
  addsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x18(%rbp)
  movl   $0xb,-0x20(%rbp)
  jmp    L3
L56:
  movsd  -0x18(%rbp),%xmm0
  movsd  D9(%rip),%xmm1
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x10(%rbp)
  movsd  -0x10(%rbp),%xmm1
  movsd  D9(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movsd  -0x18(%rbp),%xmm0
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  call   <floor@plt>
  cvttsd2si %xmm0,%eax
  mov    %eax,-0x1c(%rbp)
  cmpl   $0x9,-0x1c(%rbp)
  jg     L41
  mov    -0x1c(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edx
  jmp    L36
L41:
  mov    -0x1c(%rbp),%eax
  add    $0x57,%eax
  mov    %eax,%edx
L36:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
  movsd  -0x10(%rbp),%xmm0
  movsd  %xmm0,-0x18(%rbp)
  subl   $0x1,-0x20(%rbp)
L3:
  cmpl   $0x3,-0x20(%rbp)
  jg     L56
  mov    -0x24(%rbp),%eax
  add    $0x16356b,%eax
  mov    %eax,-0x8(%rbp)
  movl   $0x4,-0x20(%rbp)
  jmp    L18
L38:
  mov    -0x8(%rbp),%eax
  mov    $0x24,%esi
  mov    %eax,%edi
  call   <ldiv@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x4(%rbp),%eax
  cmp    $0x9,%eax
  jg     L20
  mov    -0x4(%rbp),%eax
  add    $0x30,%eax
  mov    %eax,%edx
  jmp    L63
L20:
  mov    -0x4(%rbp),%eax
  add    $0x57,%eax
  mov    %eax,%edx
L63:
  mov    -0x20(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x38(%rbp),%rax
  add    %rcx,%rax
  mov    %dl,(%rax)
L18:
  mov    -0x20(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x20(%rbp)
  test   %eax,%eax
  jne    L38
  nop
  nop
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_tag: D11
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D27
data_type: byte[8]
name: extract_symbolic_data
arguments: data_tag: D34
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D18
data_type: f64
name: extract_symbolic_data
arguments: data_tag: D9
data_type: f64

tool: D11: 1000000000

tool: {"D27": [0x00,0x00,0x00,0x08,0xba,0x86,0x84,0x42]}

tool: D34: 86400000000000

tool: D18: 0.5

tool: D9: 36

assistant: ```c
void timename(int days, int secs, int nanosecs, char *str) {
  double ntime;
  ldiv_t d;
  int i;
  ntime = (double)secs * 1E9 + (double)nanosecs;
  ntime = (ntime * (36.0 * 36.0 * 36.0 * 36.0 * 36.0 * 36.0 * 36.0 * 36.0)) / 86400E9;
  ntime = floor(ntime + 0.5);
  for (i = 11; i >= 4; i--) {
    double _d = floor(ntime / 36.0);
    int rem = (int)(floor(ntime - (_d * 36.0)));
    str[i] = rem < 10 ? '0' + rem : 'a' + rem - 10;
    ntime = _d;
  }
  d.quot = days + (25 * 365 + 6) + (31 * 36 * 36 * 36);
  for (i = 4; i--;) {
    d = ldiv(d.quot, 36);
    str[i] = d.rem < 10 ? '0' + (char)d.rem : 'a' + (char)d.rem - 10;
  }
}
```