user: 请显示与下面列出的汇编代码相匹配的C语言实现:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D14(%rip),%rax
  mov    %rax,-0x18(%rbp)
  movl   $0x6e686f4a,-0xd(%rbp)
  movb   $0x0,-0x9(%rbp)
  lea    -0xd(%rbp),%rax
  mov    %rax,-0x18(%rbp)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <puts@plt>
  mov    -0x18(%rbp),%rax
  lea    D79(%rip),%rdx
  mov    %rax,%rsi
  lea    D80(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D5(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x1,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D46(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x2,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D29(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x3,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D75(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x4,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D100(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x5,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D49(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x6,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D73(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x18(%rbp),%rax
  add    $0x7,%rax
  movzbl (%rax),%eax
  movsbl %al,%eax
  mov    %eax,%esi
  lea    D41(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

tool_calls: name: read_data
arguments: data_label: D14
datatype: string
name: read_data
arguments: data_label: D79
datatype: string
name: read_data
arguments: data_label: D80
datatype: string
name: read_data
arguments: data_label: D5
datatype: string
name: read_data
arguments: data_label: D46
datatype: string
name: read_data
arguments: data_label: D29
datatype: string
name: read_data
arguments: data_label: D75
datatype: string
name: read_data
arguments: data_label: D100
datatype: string
name: read_data
arguments: data_label: D49
datatype: string
name: read_data
arguments: data_label: D73
datatype: string
name: read_data
arguments: data_label: D41
datatype: string

tool: D14: Mugurel

tool: D79:       

tool: D80: 1st letter of that is %s%s


tool: D5: 1st letter of that is %c


tool: D46: 2nd letter of that is %c


tool: D29: 3rd letter of that is %c


tool: D75: 4th letter of that is %c


tool: D100: 5th letter of that is %c


tool: D49: 6th letter of that is %c


tool: D73: 7th letter of that is %c


tool: D41: 8th letter of that is %c



assistant: ```c
int main(void) {
  char *nume = "Mugurel";
  char aaray[] = "John";
  nume = aaray;
  puts(nume);
  printf("1st letter of that is %s\b\b\b\b\b\b%s\n", nume, "      ");
  printf("1st letter of that is %c\n", *(nume));
  printf("2nd letter of that is %c\n", *(nume + 1));
  printf("3rd letter of that is %c\n", nume[2]);
  printf("4th letter of that is %c\n", *(nume + 3));
  printf("5th letter of that is %c\n", nume[4]);
  printf("6th letter of that is %c\n", *(nume + 5));
  printf("7th letter of that is %c\n", nume[6]);
  printf("8th letter of that is %c\n\n", *(nume + 7));
  return 0;
}
```