user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<fannkuch>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x38,%rsp
  movq   $0x0,0x8(%rsp)
  test   %edi,%edi
  jg     L63
L147:
  mov    0x8(%rsp),%rax
  add    $0x38,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L63:
  mov    %edi,%r13d
  mov    $0x4,%esi
  call   <calloc@plt>
  mov    %rax,%rbx
  mov    $0x4,%esi
  mov    %r13d,%edi
  call   <calloc@plt>
  mov    %rax,%r12
  mov    $0x4,%esi
  mov    %r13d,%edi
  call   <calloc@plt>
  mov    %rax,%r14
  mov    %r13d,%edx
  mov    $0x0,%eax
L27:
  mov    %eax,(%r12,%rax,4)
  add    $0x1,%rax
  cmp    %rdx,%rax
  jne    L27
  lea    -0x1(%r13),%eax
  mov    %eax,0x1c(%rsp)
  cltq
  lea    (%r12,%rax,4),%rax
  mov    %rax,0x10(%rsp)
  mov    %r13d,%eax
  lea    (%r12,%rax,4),%rax
  mov    %rax,0x20(%rsp)
  mov    %r13d,%ebp
  movl   $0x0,0x18(%rsp)
  movq   $0x0,0x8(%rsp)
L123:
  cmpl   $0x1d,0x18(%rsp)
  jle    L225
L101:
  movslq %ebp,%rax
  cmp    $0x1,%ebp
  je     L228
L170:
  mov    %eax,-0x4(%r14,%rax,4)
  sub    $0x1,%rax
  cmp    $0x1,%eax
  jne    L170
L228:
  cmpl   $0x0,(%r12)
  je     L223
  mov    0x10(%rsp),%rax
  mov    0x1c(%rsp),%edi
  cmp    %edi,(%rax)
  jne    L86
L223:
  cmp    $0x1,%r13d
  je     L147
  lea    0x4(%r12),%rcx
  lea    0x4(%r14),%rsi
  mov    $0x1,%ebp
  jmp    L34
L225:
  mov    %r12,%r15
  mov    %rbx,0x28(%rsp)
  mov    0x20(%rsp),%rbx
L29:
  mov    (%r15),%eax
  lea    0x1(%rax),%esi
  lea    D3(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x4,%r15
  cmp    %rbx,%r15
  jne    L29
  mov    0x28(%rsp),%rbx
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0x18(%rsp)
  jmp    L101
L86:
  cmp    $0x1,%r13d
  jle    L75
  mov    %r13d,%ecx
  mov    $0x1,%eax
L88:
  mov    (%r12,%rax,4),%edx
  mov    %edx,(%rbx,%rax,4)
  add    $0x1,%rax
  cmp    %rcx,%rax
  jne    L88
L75:
  mov    (%r12),%edi
  mov    $0x0,%r8d
  jmp    L192
L189:
  mov    %edx,%edi
L192:
  lea    -0x1(%rdi),%eax
  cmp    $0x1,%eax
  jle    L59
  cltq
  mov    $0x1,%edx
L172:
  mov    (%rbx,%rdx,4),%ecx
  mov    (%rbx,%rax,4),%esi
  mov    %esi,(%rbx,%rdx,4)
  mov    %ecx,(%rbx,%rax,4)
  add    $0x1,%rdx
  sub    $0x1,%rax
  cmp    %eax,%edx
  jl     L172
L59:
  add    $0x1,%r8
  movslq %edi,%rax
  lea    (%rbx,%rax,4),%rax
  mov    (%rax),%edx
  mov    %edi,(%rax)
  test   %edx,%edx
  jne    L189
  mov    0x8(%rsp),%rax
  cmp    %r8,%rax
  cmovge %rax,%r8
  mov    %r8,0x8(%rsp)
  jmp    L223
L61:
  mov    %edi,(%rcx)
  mov    (%rsi),%eax
  sub    $0x1,%eax
  mov    %eax,(%rsi)
  test   %eax,%eax
  jg     L123
  add    $0x1,%ebp
  add    $0x4,%rcx
  add    $0x4,%rsi
  cmp    %ebp,%r13d
  je     L147
L34:
  mov    (%r12),%edi
  test   %ebp,%ebp
  jle    L61
  mov    %r12,%rax
L14:
  mov    0x4(%rax),%edx
  mov    %edx,(%rax)
  add    $0x4,%rax
  cmp    %rcx,%rax
  jne    L14
  jmp    L61
```

tool_calls: name: unpack_data
arguments: data_label: D3
data_type: string

tool: D3: %d

assistant: ```c
long fannkuch(int n) {
  int *perm;
  int *perm1;
  int *count;
  long flips;
  long flipsMax;
  int r;
  int i;
  int k;
  int didpr;
  const int n1 = n - 1;
  if (n < 1)
    return 0;
  perm = calloc(n, sizeof(*perm));
  perm1 = calloc(n, sizeof(*perm1));
  count = calloc(n, sizeof(*count));
  for (i = 0; i < n; ++i)
    perm1[i] = i;
  r = n;
  didpr = 0;
  flipsMax = 0;
  for (;;) {
    if (didpr < 30) {
      for (i = 0; i < n; ++i)
        printf("%d", (int)(1 + perm1[i]));
      printf("\n");
      ++didpr;
    }
    for (; r != 1; --r) {
      count[r - 1] = r;
    }
    if (!(perm1[0] == 0 || perm1[n1] == n1)) {
      flips = 0;
      for (i = 1; i < n; ++i) {
        perm[i] = perm1[i];
      }
      k = perm1[0];
      do {
        int j;
        for (i = 1, j = k - 1; i < j; ++i, --j) {
          {
            int t_mp;
            t_mp = (perm[i]);
            (perm[i]) = (perm[j]);
            (perm[j]) = t_mp;
          }
        }
        ++flips;
        j = perm[k];
        perm[k] = k;
        k = j;
      } while (k);
      if (flipsMax < flips) {
        flipsMax = flips;
      }
    }
    for (;;) {
      if (r == n) {
        return flipsMax;
      }
      {
        int perm0 = perm1[0];
        i = 0;
        while (i < r) {
          k = i + 1;
          perm1[i] = perm1[k];
          i = k;
        }
        perm1[r] = perm0;
      }
      if ((count[r] -= 1) > 0) {
        break;
      }
      ++r;
    }
  }
}
```