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   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbx
  sub    $0x28,%rsp
  mov    %edi,%r14d
  mov    %esi,-0x44(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  movslq %esi,%rdx
  lea    0xf(%rdx),%rax
  mov    %rax,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rcx
  sub    %rax,%rcx
L88:
  cmp    %rcx,%rsp
  je     L86
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L88
L86:
  mov    %rsi,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L95
  orq    $0x0,-0x8(%rsp,%rax,1)
L95:
  mov    %rsp,%r12
  lea    0xf(,%rdx,8),%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L66:
  cmp    %rdx,%rsp
  je     L108
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L66
L108:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L36
  orq    $0x0,-0x8(%rsp,%rax,1)
L36:
  mov    %rsp,%r13
  mov    -0x44(%rbp),%eax
  test   %eax,%eax
  jle    L42
  mov    %eax,%r15d
  mov    $0x0,%ebx
L39:
  mov    $0x0,%eax
  call   <rand_double@plt>
  movsd  D8(%rip),%xmm1
  comisd %xmm0,%xmm1
  seta   (%r12,%rbx,1)
  mov    $0x0,%eax
  call   <rand_double@plt>
  mulsd  D25(%rip),%xmm0
  subsd  D33(%rip),%xmm0
  movsd  %xmm0,0x0(%r13,%rbx,8)
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L39
L42:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%r15
  test   %r14d,%r14d
  jle    L73
  mov    $0x0,%ebx
L96:
  mov    $0x1,%r8d
  mov    %r12,%rcx
  mov    $0x1,%edx
  mov    %r13,%rsi
  mov    -0x44(%rbp),%edi
  call   <stdlib_strided_dmskmin@plt>
  ucomisd %xmm0,%xmm0
  jp     L91
  add    $0x1,%ebx
  cmp    %ebx,%r14d
  jne    L96
L73:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r15,%xmm2
  subsd  %xmm2,%xmm0
  movq   %xmm0,%rbx
L120:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L99
  movq   %rbx,%xmm0
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L91:
  lea    D39(%rip),%r12
  mov    %r12,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r15,%xmm3
  subsd  %xmm3,%xmm0
  movq   %xmm0,%rbx
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L120
L99:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: tag: D8
datatype: f64
name: unpack_data
arguments: tag: D25
datatype: f64
name: unpack_data
arguments: tag: D33
datatype: f64
name: unpack_data
arguments: tag: D39
datatype: string

tool: D8: 0.2

tool: D25: 20000

tool: D33: 10000

tool: D39: should not return NaN


assistant: ```c
double benchmark(int iterations, int len) {
  unsigned char mask[len];
  double elapsed;
  double x[len];
  double v;
  double t;
  int i;
  for (i = 0; i < len; i++) {
    if (rand_double() < 0.2) {
      mask[i] = 1;
    } else {
      mask[i] = 0;
    }
    x[i] = (rand_double() * 20000.0) - 10000.0;
  }
  v = 0.0;
  t = tic();
  for (i = 0; i < iterations; i++) {
    v = stdlib_strided_dmskmin(len, x, 1, mask, 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;
}
```