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
  movslq %esi,%r13
  push   %r12
  push   %rbx
  mov    %r13,%rbx
  sub    $0x28,%rsp
  mov    %edi,-0x48(%rbp)
  mov    %rsp,%rcx
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  lea    0xf(,%r13,4),%rax
  mov    %rax,%rsi
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rsi
  and    $0xfffffffffffffff0,%rdx
  sub    %rsi,%rcx
  cmp    %rcx,%rsp
  je     L123
L87:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L87
L123:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L116
L111:
  mov    %rax,%rdx
  mov    %rsp,%rcx
  and    $0xfffffffffffff000,%rax
  mov    %rsp,%r15
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L83
L8:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L8
L83:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L101
L86:
  mov    %rsp,%r12
  test   %ebx,%ebx
  jle    L42
  xor    %r14d,%r14d
  nopl   0x0(%rax)
L76:
  xor    %eax,%eax
  call   <rand_float@plt>
  mulss  D21(%rip),%xmm0
  movl   $0x0,(%r12,%r14,4)
  subss  D17(%rip),%xmm0
  movss  %xmm0,(%r15,%r14,4)
  add    $0x1,%r14
  cmp    %r14,%r13
  jne    L76
L42:
  xor    %eax,%eax
  call   <tic@plt>
  mov    -0x48(%rbp),%eax
  movsd  %xmm0,-0x50(%rbp)
  test   %eax,%eax
  jle    L53
  xor    %r13d,%r13d
  jmp    L106
  xchg   %ax,%ax
L22:
  add    $0x1,%r13d
  cmp    %r13d,-0x48(%rbp)
  je     L53
L106:
  movss  D22(%rip),%xmm0
  mov    %r12,%rcx
  mov    %r15,%rsi
  mov    %ebx,%edi
  addss  (%r15),%xmm0
  mov    $0x1,%r8d
  mov    $0x1,%edx
  movss  %xmm0,(%r15)
  pxor   %xmm0,%xmm0
  call   <stdlib_strided_scusumpw@plt>
  movss  (%r12),%xmm0
  ucomiss %xmm0,%xmm0
  jnp    L22
  lea    D30(%rip),%rdi
  call   <printf@plt>
L53:
  sub    $0x1,%ebx
  xor    %eax,%eax
  call   <tic@plt>
  movslq %ebx,%rbx
  subsd  -0x50(%rbp),%xmm0
  movss  (%r12,%rbx,4),%xmm1
  ucomiss %xmm1,%xmm1
  jp     L135
L75:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L11
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  nopw   0x0(%rax,%rax,1)
L116:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L111
  nopl   0x0(%rax,%rax,1)
L135:
  lea    D30(%rip),%rdi
  movsd  %xmm0,-0x48(%rbp)
  call   <printf@plt>
  movsd  -0x48(%rbp),%xmm0
  jmp    L75
L101:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L86
L11:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: tag: D21
type_guessed: byte[4]
name: retrieve_data
arguments: tag: D17
type_guessed: byte[4]
name: retrieve_data
arguments: tag: D22
type_guessed: byte[4]
name: retrieve_data
arguments: tag: D30
type_guessed: string

tool: {"D21": [0x00,0x40,0x9c,0x46]}

tool: {"D17": [0x00,0x40,0x1c,0x46]}

tool: {"D22": [0x00,0x00,0x80,0x3f]}

tool: D30: should not return NaN


assistant: ```c
double benchmark(int iterations, int len) {
  double elapsed;
  float x[len];
  float y[len];
  double t;
  int i;
  for (i = 0; i < len; i++) {
    x[i] = (rand_float() * 20000.0f) - 10000.0f;
    y[i] = 0.0f;
  }
  t = tic();
  for (i = 0; i < iterations; i++) {
    x[0] += 1.0f;
    stdlib_strided_scusumpw(len, 0.0f, x, 1, y, 1);
    if (y[0] != y[0]) {
      printf("should not return NaN\n");
      break;
    }
  }
  elapsed = tic() - t;
  if (y[len - 1] != y[len - 1]) {
    printf("should not return NaN\n");
  }
  return elapsed;
}
```