user: 请从给定的汇编指令推导出C源代码:

```asm
<benchmark>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x58,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %esi,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  mov    -0x58(%rbp),%eax
  movslq %eax,%rdx
  sub    $0x1,%rdx
  mov    %rdx,-0x38(%rbp)
  movslq %eax,%rdx
  mov    %rdx,%r10
  mov    $0x0,%r11d
  movslq %eax,%rdx
  mov    %rdx,%r8
  mov    $0x0,%r9d
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%esi
  mov    $0x0,%edx
  div    %rsi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L54:
  cmp    %rdx,%rsp
  je     L87
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L54
L87:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L48
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L48:
  mov    %rsp,%rax
  add    $0x7,%rax
  shr    $0x3,%rax
  shl    $0x3,%rax
  mov    %rax,-0x30(%rbp)
  movl   $0x0,-0x44(%rbp)
  jmp    L12
L7:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D22(%rip),%xmm1
  mulsd  %xmm1,%xmm0
  movsd  D33(%rip),%xmm1
  subsd  %xmm1,%xmm0
  mov    -0x30(%rbp),%rax
  mov    -0x44(%rbp),%edx
  movslq %edx,%rdx
  movsd  %xmm0,(%rax,%rdx,8)
  addl   $0x1,-0x44(%rbp)
L12:
  mov    -0x44(%rbp),%eax
  cmp    -0x58(%rbp),%eax
  jl     L7
  pxor   %xmm0,%xmm0
  movsd  %xmm0,-0x40(%rbp)
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x28(%rbp)
  movl   $0x0,-0x44(%rbp)
  jmp    L47
L83:
  mov    -0x30(%rbp),%rdx
  mov    -0x58(%rbp),%eax
  mov    $0x1,%ecx
  mov    $0x1,%esi
  mov    %eax,%edi
  call   <stdlib_strided_dsemtk@plt>
  movq   %xmm0,%rax
  mov    %rax,-0x40(%rbp)
  movsd  -0x40(%rbp),%xmm0
  ucomisd -0x40(%rbp),%xmm0
  jnp    L35
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L61
L35:
  addl   $0x1,-0x44(%rbp)
L47:
  mov    -0x44(%rbp),%eax
  cmp    -0x54(%rbp),%eax
  jl     L83
L61:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%rax
  movq   %rax,%xmm0
  subsd  -0x28(%rbp),%xmm0
  movsd  %xmm0,-0x20(%rbp)
  movsd  -0x40(%rbp),%xmm0
  ucomisd -0x40(%rbp),%xmm0
  jnp    L93
  lea    D37(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L93:
  movsd  -0x20(%rbp),%xmm0
  mov    %rbx,%rsp
  movq   %xmm0,%rax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L28
  call   <__stack_chk_fail@plt>
L28:
  movq   %rax,%xmm0
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: tag: D22
type_guessed: f64
name: unpack_data
arguments: tag: D33
type_guessed: f64
name: unpack_data
arguments: tag: D37
type_guessed: string

tool: D22: 20000

tool: D33: 10000

tool: D37: should not return NaN


assistant: ```c
double benchmark(int iterations, int len) {
  double elapsed;
  double x[len];
  double v;
  double t;
  int i;
  for (i = 0; i < len; i++) {
    x[i] = (rand_double() * 20000.0) - 10000.0;
  }
  v = 0.0;
  t = tic();
  for (i = 0; i < iterations; i++) {
    v = stdlib_strided_dsemtk(len, 1, x, 1);
    if (v != v) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (v != v) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```