user: 这段汇编代码用C语言表示是什么样的？

```asm
<random_bench>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %edi,%r12d
  mov    %esi,%ebp
  lea    D6(%rip),%rbx
  lea    0x40000(%rbx),%r13
L18:
  mov    $0x0,%eax
  call   <rand@plt>
  movslq %eax,%rdx
  imul   $0x666665c3,%rdx,%rdx
  sar    $0x36,%rdx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  imul   $0xa00001,%edx,%edx
  sub    %edx,%eax
  mov    %eax,(%rbx)
  add    $0x4,%rbx
  cmp    %r13,%rbx
  jne    L18
  lea    D52(%rip),%rbx
  lea    0x20000(%rbx),%r13
L11:
  mov    $0x0,%eax
  call   <rand@plt>
  cltd
  idiv   %r12d
  add    $0x1,%edx
  mov    %edx,(%rbx)
  add    $0x4,%rbx
  cmp    %rbx,%r13
  jne    L11
  mov    $0x64,%edi
  call   <sleepms@plt>
  mov    $0x0,%eax
  call   <gettime@plt>
  mov    %eax,0x8(%rsp)
  test   %ebp,%ebp
  je     L46
  mov    $0x0,%ebx
  lea    D52(%rip),%r15
  lea    D6(%rip),%r12
  lea    D7(%rip),%r14
  lea    D62(%rip),%r13
L55:
  lea    (%rbx,%rbx,1),%eax
  mov    %ebx,%edx
  add    $0x1,%ebx
  and    $0x7fff,%edx
  mov    (%r15,%rdx,4),%ecx
  lea    0x1(%rax),%edx
  movzwl %dx,%edx
  movslq (%r12,%rdx,4),%rsi
  add    %r14,%rsi
  and    $0xfffe,%eax
  movslq (%r12,%rax,4),%rdi
  add    %r13,%rdi
  mov    %ecx,%edx
  call   <memcpy@plt>
  cmp    %ebp,%ebx
  jne    L55
  mov    $0x0,%eax
  call   <gettime@plt>
  sub    0x8(%rsp),%eax
  mov    %eax,0x8(%rsp)
  mov    $0x64,%edi
  call   <sleepms@plt>
  mov    $0x0,%eax
  call   <gettime@plt>
  mov    %eax,0xc(%rsp)
  mov    $0x0,%ebx
  lea    D52(%rip),%r15
  lea    D6(%rip),%r12
  lea    D7(%rip),%r14
  lea    D62(%rip),%r13
L43:
  lea    (%rbx,%rbx,1),%eax
  mov    %ebx,%edx
  add    $0x1,%ebx
  and    $0x7fff,%edx
  mov    (%r15,%rdx,4),%ecx
  lea    0x1(%rax),%edx
  movzwl %dx,%edx
  movslq (%r12,%rdx,4),%rsi
  add    %r14,%rsi
  and    $0xfffe,%eax
  movslq (%r12,%rax,4),%rdi
  add    %r13,%rdi
  mov    %ecx,%edx
  call   <memcpy_fast@plt>
  cmp    %ebp,%ebx
  jne    L43
L57:
  mov    $0x0,%eax
  call   <gettime@plt>
  mov    %eax,%ebx
  lea    D29(%rip),%rdi
  call   <puts@plt>
  mov    %ebx,%esi
  sub    0xc(%rsp),%esi
  mov    0x8(%rsp),%edx
  lea    D69(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L46:
  mov    $0x0,%eax
  call   <gettime@plt>
  sub    0x8(%rsp),%eax
  mov    %eax,0x8(%rsp)
  mov    $0x64,%edi
  call   <sleepms@plt>
  mov    $0x0,%eax
  call   <gettime@plt>
  mov    %eax,0xc(%rsp)
  jmp    L57
```

tool_calls: name: unpack_data
arguments: data_tag: D29
type: string
name: unpack_data
arguments: data_tag: D69
type: string

tool: D29: benchmark random access:

tool: D69: memcpy_fast=%dms memcpy=%dms



assistant: ```c
void random_bench(int maxsize, int times) {
  static char A[11 * 1024 * 1024 + 2];
  static char B[11 * 1024 * 1024 + 2];
  static int random_offsets[0x10000];
  static int random_sizes[0x8000];
  unsigned int i, p1, p2;
  unsigned int t1, t2;
  for (i = 0; i < 0x10000; i++) {
    random_offsets[i] = rand() % (10 * 1024 * 1024 + 1);
  }
  for (i = 0; i < 0x8000; i++) {
    random_sizes[i] = 1 + rand() % maxsize;
  }
  sleepms(100);
  t1 = gettime();
  for (p1 = 0, p2 = 0, i = 0; i < times; i++) {
    int offset1 = random_offsets[(p1++) & 0xffff];
    int offset2 = random_offsets[(p1++) & 0xffff];
    int size = random_sizes[(p2++) & 0x7fff];
    memcpy(A + offset1, B + offset2, size);
  }
  t1 = gettime() - t1;
  sleepms(100);
  t2 = gettime();
  for (p1 = 0, p2 = 0, i = 0; i < times; i++) {
    int offset1 = random_offsets[(p1++) & 0xffff];
    int offset2 = random_offsets[(p1++) & 0xffff];
    int size = random_sizes[(p2++) & 0x7fff];
    memcpy_fast(A + offset1, B + offset2, size);
  }
  t2 = gettime() - t2;
  printf("benchmark random access:\n");
  printf("memcpy_fast=%dms memcpy=%dms\n\n", (int)t2, (int)t1);
}
```