user: 请从给定的汇编指令推导出C源代码：

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  mov    %rsp,%rax
  mov    %rax,%rbx
  movl   $0x9,-0x38(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  sub    $0x1,%rax
  mov    %rax,-0x30(%rbp)
  mov    -0x38(%rbp),%eax
  cltq
  mov    %rax,%r8
  mov    $0x0,%r9d
  mov    -0x38(%rbp),%eax
  cltq
  mov    %rax,%rsi
  mov    $0x0,%edi
  mov    -0x38(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    $0x10,%eax
  sub    $0x1,%rax
  add    %rdx,%rax
  mov    $0x10,%edi
  mov    $0x0,%edx
  div    %rdi
  imul   $0x10,%rax,%rax
  mov    %rax,%rcx
  and    $0xfffffffffffff000,%rcx
  mov    %rsp,%rdx
  sub    %rcx,%rdx
L31:
  cmp    %rdx,%rsp
  je     L58
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  jmp    L31
L58:
  mov    %rax,%rdx
  and    $0xfff,%edx
  sub    %rdx,%rsp
  mov    %rax,%rdx
  and    $0xfff,%edx
  test   %rdx,%rdx
  je     L35
  and    $0xfff,%eax
  sub    $0x8,%rax
  add    %rsp,%rax
  orq    $0x0,(%rax)
L35:
  mov    %rsp,%rax
  add    $0x3,%rax
  shr    $0x2,%rax
  shl    $0x2,%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x28(%rbp),%rax
  sub    $0x4,%rax
  mov    %rax,-0x20(%rbp)
  movl   $0x0,-0x44(%rbp)
  jmp    L25
L55:
  movl   $0x9,-0x34(%rbp)
  mov    -0x34(%rbp),%eax
  imul   -0x44(%rbp),%eax
  mov    %eax,%ecx
  mov    -0x28(%rbp),%rax
  mov    -0x44(%rbp),%edx
  movslq %edx,%rdx
  mov    %ecx,(%rax,%rdx,4)
  addl   $0x1,-0x44(%rbp)
L25:
  mov    -0x44(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L55
  movl   $0x0,-0x40(%rbp)
  jmp    L82
L7:
  mov    -0x28(%rbp),%rax
  mov    -0x40(%rbp),%edx
  movslq %edx,%rdx
  mov    (%rax,%rdx,4),%edx
  mov    -0x40(%rbp),%eax
  mov    %eax,%esi
  lea    D19(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x40(%rbp)
L82:
  mov    -0x40(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jl     L7
  mov    $0xa,%edi
  call   <putchar@plt>
  movl   $0x1,-0x3c(%rbp)
  jmp    L12
L27:
  mov    -0x3c(%rbp),%eax
  cltq
  lea    0x0(,%rax,4),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%edx
  mov    -0x3c(%rbp),%eax
  mov    %eax,%esi
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x3c(%rbp)
L12:
  mov    -0x3c(%rbp),%eax
  cmp    -0x38(%rbp),%eax
  jle    L27
  mov    $0x0,%eax
  mov    %rbx,%rsp
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L76
  call   <__stack_chk_fail@plt>
L76:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: extract_symbolic_data
arguments: data_label: D19
type_guessed: string
name: extract_symbolic_data
arguments: data_label: D0
type_guessed: string

tool: D19: start_at_zero[%d] = %d


tool: D0: start_at_one[%d] = %d


assistant: ```c
int main(void) {
  const int N = 9;
  int start_at_zero[N];
  int *start_at_one = start_at_zero - 1;
  for (int i = 0; i < N; i++) {
    const int mul = 9;
    start_at_zero[i] = mul * i;
  }
  for (int i = 0; i < N; i++) {
    printf("start_at_zero[%d] = %d\n", i, start_at_zero[i]);
  }
  printf("\n");
  for (int i = 1; i <= N; i++) {
    printf("start_at_one[%d] = %d\n", i, start_at_one[i]);
  }
  return 0;
}
```