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
  mov    %esi,%eax
  cltq
  shl    $0x2,%rax
  lea    0xf(%rax),%rdx
  mov    %rdx,%rsi
  and    $0xfffffffffffffff0,%rsi
  and    $0xfffffffffffff000,%rdx
  mov    %rsp,%rcx
  sub    %rdx,%rcx
L102:
  cmp    %rcx,%rsp
  je     L65
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L102
L65:
  mov    %rsi,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  je     L53
  orq    $0x0,-0x8(%rsp,%rdx,1)
L53:
  mov    %rsp,%r12
  add    $0xf,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffffff0,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%rdx
  sub    %rax,%rdx
L15:
  cmp    %rdx,%rsp
  je     L24
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L15
L24:
  mov    %rcx,%rax
  and    $0xfff,%eax
  sub    %rax,%rsp
  test   %rax,%rax
  je     L37
  orq    $0x0,-0x8(%rsp,%rax,1)
L37:
  mov    %rsp,%r13
  mov    -0x44(%rbp),%eax
  test   %eax,%eax
  jle    L91
  mov    %eax,%r15d
  mov    $0x0,%ebx
L87:
  mov    $0x0,%eax
  call   <rand_float@plt>
  mulss  D20(%rip),%xmm0
  subss  D28(%rip),%xmm0
  movss  %xmm0,(%r12,%rbx,4)
  mov    $0x0,%eax
  call   <rand_float@plt>
  mulss  D20(%rip),%xmm0
  subss  D28(%rip),%xmm0
  movss  %xmm0,0x0(%r13,%rbx,4)
  add    $0x1,%rbx
  cmp    %r15,%rbx
  jne    L87
L91:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %xmm0,%r15
  test   %r14d,%r14d
  jle    L3
  mov    $0x0,%ebx
L48:
  mov    $0x1,%r8d
  mov    %r13,%rcx
  mov    $0x1,%edx
  mov    %r12,%rsi
  mov    -0x44(%rbp),%edi
  call   <c_dsdot@plt>
  ucomisd %xmm0,%xmm0
  jp     L85
  add    $0x1,%ebx
  cmp    %ebx,%r14d
  jne    L48
L3:
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r15,%xmm1
  subsd  %xmm1,%xmm0
  movq   %xmm0,%rbx
L7:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L83
  movq   %rbx,%xmm0
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
L85:
  lea    D3(%rip),%r12
  mov    %r12,%rdi
  call   <printf@plt>
  mov    $0x0,%eax
  call   <tic@plt>
  movq   %r15,%xmm2
  subsd  %xmm2,%xmm0
  movq   %xmm0,%rbx
  mov    %r12,%rdi
  call   <printf@plt>
  jmp    L7
L83:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: data_label: D20
type_infered: byte[4]
name: parse_data
arguments: data_label: D28
type_infered: byte[4]
name: parse_data
arguments: data_label: D3
type_infered: string

tool: {"D20": [0x00,0x40,0x9c,0x46]}

tool: {"D28": [0x00,0x40,0x1c,0x46]}

tool: D3: should not return NaN


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