user: 这段汇编代码用C语言表示是什么样的:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  add    $0xffffffffffffff80,%rsp
  mov    %edi,-0x74(%rbp)
  mov    %rsi,-0x80(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    -0x50(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  mov    $0x0,%eax
  call   <yogrt_remaining@plt>
  mov    %eax,-0x6c(%rbp)
  lea    -0x30(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movl   $0x0,-0x70(%rbp)
  jmp    L0
L23:
  mov    $0x0,%eax
  call   <yogrt_remaining@plt>
  mov    %eax,-0x6c(%rbp)
  addl   $0x1,-0x70(%rbp)
L0:
  cmpl   $0x270f,-0x70(%rbp)
  jle    L23
  lea    -0x40(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <gettimeofday@plt>
  movsd  -0x50(%rbp),%xmm1
  movsd  -0x48(%rbp),%xmm2
  movsd  D24(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x68(%rbp)
  movsd  -0x30(%rbp),%xmm1
  movsd  -0x28(%rbp),%xmm2
  movsd  D24(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x60(%rbp)
  movsd  -0x40(%rbp),%xmm1
  movsd  -0x38(%rbp),%xmm2
  movsd  D24(%rip),%xmm0
  mulsd  %xmm2,%xmm0
  addsd  %xmm1,%xmm0
  movsd  %xmm0,-0x58(%rbp)
  movsd  -0x60(%rbp),%xmm0
  movapd %xmm0,%xmm1
  subsd  -0x68(%rbp),%xmm1
  movsd  D12(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D70(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  lea    D89(%rip),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  movsd  -0x58(%rbp),%xmm0
  movapd %xmm0,%xmm1
  subsd  -0x60(%rbp),%xmm1
  movsd  D3(%rip),%xmm0
  divsd  %xmm1,%xmm0
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  movsd  -0x58(%rbp),%xmm0
  subsd  -0x60(%rbp),%xmm0
  movsd  D3(%rip),%xmm2
  movapd %xmm0,%xmm1
  divsd  %xmm2,%xmm1
  movsd  D47(%rip),%xmm0
  mulsd  %xmm0,%xmm1
  movq   %xmm1,%rax
  movq   %rax,%xmm0
  lea    D68(%rip),%rax
  mov    %rax,%rdi
  mov    $0x1,%eax
  call   <printf@plt>
  mov    $0x0,%edi
  call   <exit@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L7
  call   <__stack_chk_fail@plt>
L7:
  leave
  ret
```

tool_calls: name: parse_data
arguments: label: D24
data_type: f64
name: parse_data
arguments: label: D12
data_type: byte[8]
name: parse_data
arguments: label: D70
data_type: string
name: parse_data
arguments: label: D89
data_type: string
name: parse_data
arguments: label: D3
data_type: byte[8]
name: parse_data
arguments: label: D53
data_type: string
name: parse_data
arguments: label: D47
data_type: byte[8]
name: parse_data
arguments: label: D68
data_type: string

tool: D24: 1e-06

tool: {"D12": [0x00,0x00,0x00,0x00,0x00,0x40,0x8f,0x40]}

tool: D70: yogrt_remaining first call:
	%.03f msec


tool: D89: yogrt_remaining cached timing:

tool: {"D3": [0x00,0x00,0x00,0x00,0x00,0x88,0xc3,0x40]}

tool: D53: 	%.0f calls/sec


tool: {"D47": [0x00,0x00,0x00,0x00,0x00,0x6a,0xf8,0x40]}

tool: D68: 	%.3f usecs/call


assistant: ```c
main(int argc, char *argv[]) {
  int i;
  int rem;
  struct timeval tv1[1], tv2[2], tv3[1];
  double t1, t2, t3;
  gettimeofday(tv1, NULL);
  rem = yogrt_remaining();
  gettimeofday(tv2, NULL);
  for (i = 0; i < 10000; i++) {
    rem = yogrt_remaining();
  }
  gettimeofday(tv3, NULL);
  t1 = tv1->tv_sec + (tv1->tv_usec * (double)0.000001);
  t2 = tv2->tv_sec + (tv2->tv_usec * (double)0.000001);
  t3 = tv3->tv_sec + (tv3->tv_usec * (double)0.000001);
  printf("yogrt_remaining first call:\n\t%.03f msec\n", (t2 - t1) * 1000);
  printf("yogrt_remaining cached timing:\n");
  printf("\t%.0f calls/sec\n", 10000 / (t3 - t2));
  printf("\t%.3f usecs/call\n", (t3 - t2) / 10000 * 100000);
  exit(0);
}
```