user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r14
  mov    $0x6f,%edx
  lea    D9(%rip),%rdi
  push   %r13
  lea    D96(%rip),%r13
  push   %r12
  lea    D110(%rip),%r12
  push   %rbp
  push   %rbx
  lea    D13(%rip),%rbx
  mov    %rbx,%rsi
  lea    0x6(%rbx),%r14
  sub    $0x10,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  lea    0x2(%rsp),%rbp
  mov    %dx,0x6(%rsp)
  movl   $0x6c6c6568,0x2(%rsp)
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D124(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D103(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D65(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D78(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbp,%rsi
  lea    D18(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x8,%esi
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    $0x6,%esi
  lea    D112(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  movsbl 0x2(%rsp),%esi
  mov    %r12,%rdi
  mov    %rbp,%rdx
  xor    %eax,%eax
  mov    %esi,%ecx
  call   <printf@plt>
  mov    %r12,%rdi
  mov    $0x48,%ecx
  mov    %rbp,%rdx
  mov    $0x48,%esi
  xor    %eax,%eax
  movb   $0x48,0x2(%rsp)
  call   <printf@plt>
  mov    %r12,%rdi
  mov    $0x68,%ecx
  mov    %rbx,%rdx
  mov    $0x68,%esi
  xor    %eax,%eax
  lea    D51(%rip),%r12
  call   <printf@plt>
  mov    $0x68,%ecx
  xor    %eax,%eax
  mov    $0x68,%edx
  mov    $0x68,%esi
  lea    D102(%rip),%rdi
  call   <printf@plt>
  nopl   (%rax)
L14:
  mov    %rbx,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r12,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  add    $0x1,%rbp
  cmp    %r14,%rbx
  jne    L14
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L0
  add    $0x10,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  ret
L0:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: extract_symbolic_data
arguments: label: D9
type: string
name: extract_symbolic_data
arguments: label: D96
type: string
name: extract_symbolic_data
arguments: label: D110
type: string
name: extract_symbolic_data
arguments: label: D13
type: string
name: extract_symbolic_data
arguments: label: D124
type: string
name: extract_symbolic_data
arguments: label: D103
type: string
name: extract_symbolic_data
arguments: label: D65
type: string
name: extract_symbolic_data
arguments: label: D78
type: string
name: extract_symbolic_data
arguments: label: D18
type: string
name: extract_symbolic_data
arguments: label: D58
type: string
name: extract_symbolic_data
arguments: label: D112
type: string
name: extract_symbolic_data
arguments: label: D51
type: string
name: extract_symbolic_data
arguments: label: D102
type: string

tool: D9: s1=[%s]


tool: D96: s1: [%s] 


tool: D110: %c 	 %20i 	 %c 


tool: D13: hello

tool: D124: s2=[%s]


tool: D103: s3=[%s]


tool: D65: s4=[%s]


tool: D78: s5=[%s]


tool: D18: s6=[%s]


tool: D58: size of s1: %d 


tool: D112: size of s2: %d 


tool: D51: s2: [%s] 


tool: D102: %c 	 %20c 	 %c 


assistant: ```c
int main() {
  char *s1 = "hello";
  char s2[] = "hello";
  char *s3 = s2;
  char **s4 = &s3;
  char **s5 = &s1;
  char(*s6)[6] = &s2;
  printf("s1=[%s]\n", s1);
  printf("s2=[%s]\n", s2);
  printf("s3=[%s]\n", s3);
  printf("s4=[%s]\n", *s4);
  printf("s5=[%s]\n", *s5);
  printf("s6=[%s]\n", s6);
  printf("size of s1: %d \n", sizeof(s1));
  printf("size of s2: %d \n", sizeof(s2));
  printf("%c \t %20i \t %c \n", s3[0], s6[0], *(s6[0]));
  s2[0] = 'H';
  printf("%c \t %20i \t %c \n", s3[0], s6[0], (*s6)[0]);
  printf("%c \t %20i \t %c \n", s1[0], s5[0], *s5[0]);
  printf("%c \t %20c \t %c \n", s1[0], (*s5)[0], *(s5[0]));
  for (int i = 0; i < 6; i++) {
    printf("s1: [%s] \n", s1 + i);
    printf("s2: [%s] \n", s2 + i);
  }
  return 0;
}
```