user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<enable_prefetch_core2>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %edi,%r12d
  lea    D22(%rip),%rdi
  call   <puts@plt>
  mov    %r12d,%r14d
  cmp    $0xffffffff,%r12d
  je     L36
L37:
  movabs $0x8000000000,%r13
  lea    D6(%rip),%r15
L3:
  mov    %r12d,%edi
  call   <open_msr@plt>
  mov    %eax,%ebp
  test   %eax,%eax
  js     L6
  mov    $0x1a0,%esi
  mov    %eax,%edi
  call   <read_msr@plt>
  mov    %rax,%rbx
  and    %r13,%rax
  cmp    $0x1,%rax
  sbb    %esi,%esi
  and    $0xb,%esi
  movabs $0x2000000000,%rax
  and    %rbx,%rax
  cmp    $0x1,%rax
  sbb    %eax,%eax
  and    $0xb,%eax
  mov    %rbx,%rdx
  and    $0x80000,%edx
  cmp    $0x1,%rdx
  sbb    %ecx,%ecx
  and    $0xb,%ecx
  add    $0x4e,%ecx
  mov    %rbx,%rdx
  and    $0x200,%edx
  cmp    $0x1,%rdx
  sbb    %edx,%edx
  and    $0xb,%edx
  add    $0x4e,%edx
  lea    0x4e(%rsi),%r9d
  lea    0x4e(%rax),%r8d
  mov    %r12d,%esi
  mov    %r15,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movabs $0xffffff5ffff7fdff,%rdx
  and    %rbx,%rdx
  mov    $0x1a0,%esi
  mov    %ebp,%edi
  call   <write_msr@plt>
  mov    $0x1a0,%esi
  mov    %ebp,%edi
  call   <read_msr@plt>
  mov    %rax,%rdx
  and    %r13,%rdx
  cmp    $0x1,%rdx
  sbb    %edi,%edi
  and    $0xb,%edi
  movabs $0x2000000000,%rdx
  and    %rax,%rdx
  cmp    $0x1,%rdx
  sbb    %esi,%esi
  and    $0xb,%esi
  mov    %rax,%rdx
  and    $0x80000,%edx
  cmp    $0x1,%rdx
  sbb    %ecx,%ecx
  and    $0xb,%ecx
  add    $0x4e,%ecx
  and    $0x200,%eax
  cmp    $0x1,%rax
  sbb    %edx,%edx
  and    $0xb,%edx
  add    $0x4e,%edx
  lea    0x4e(%rdi),%r9d
  lea    0x4e(%rsi),%r8d
  mov    %r12d,%esi
  lea    D9(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    %ebp,%edi
  call   <close@plt>
  add    $0x1,%r12d
  cmp    %r12d,%r14d
  jge    L3
L6:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L36:
  mov    $0x400,%r14d
  mov    $0x0,%r12d
  jmp    L37
```

tool_calls: name: unpack_data
arguments: tag: D22
type: string
name: unpack_data
arguments: tag: D6
type: string
name: unpack_data
arguments: tag: D9
type: string

tool: D22: Enable all prefetch

tool: D6: 	Core %d old : L2HW=%c L2ADJ=%c DCU=%c DCUIP=%c


tool: D9: 	Core %d new : L2HW=%c L2ADJ=%c DCU=%c DCUIP=%c


assistant: ```c
int enable_prefetch_core2(int core) {
  int fd;
  uint64_t result;
  int begin, end, c;
  printf("Enable all prefetch\n");
  if (core == -1) {
    begin = 0;
    end = 1024;
  } else {
    begin = core;
    end = core;
  }
  for (c = begin; c <= end; c++) {
    fd = open_msr(c);
    if (fd < 0)
      break;
    result = read_msr(fd, 0x1a0);
    printf("\tCore %d old : L2HW=%c L2ADJ=%c DCU=%c DCUIP=%c\n", c, result & (1ULL << 9) ? 'N' : 'Y', result & (1ULL << 19) ? 'N' : 'Y', result & (1ULL << 37) ? 'N' : 'Y', result & (1ULL << 39) ? 'N' : 'Y');
    result &= ~((1ULL << 9) | (1ULL << 19) | (1ULL << 37) | (1ULL << 39));
    result = write_msr(fd, 0x1a0, result);
    result = read_msr(fd, 0x1a0);
    printf("\tCore %d new : L2HW=%c L2ADJ=%c DCU=%c DCUIP=%c\n", c, result & (1ULL << 9) ? 'N' : 'Y', result & (1ULL << 19) ? 'N' : 'Y', result & (1ULL << 37) ? 'N' : 'Y', result & (1ULL << 39) ? 'N' : 'Y');
    close(fd);
  }
  return 0;
}
```