user: 请从给定的汇编指令推导出C源代码:

```asm
<main>:
  endbr64
  push   %r13
  lea    D84(%rip),%rdi
  push   %r12
  lea    D58(%rip),%r12
  push   %rbp
  xor    %ebp,%ebp
  push   %rbx
  sub    $0x48,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x38(%rsp)
  movabs $0x4443424132727473,%rax
  lea    0x10(%rsp),%rbx
  mov    %rax,0x6(%rsp)
  mov    $0x45,%eax
  mov    %ax,0xe(%rsp)
  xor    %eax,%eax
  call   <printf@plt>
  mov    %rbx,%rsi
  lea    D15(%rip),%rdi
  call   <scanf@plt>
  mov    %rbx,%rsi
  lea    D81(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L36
  nopl   0x0(%rax,%rax,1)
L13:
  add    $0x1,%ebp
  add    $0x1,%rbx
  cmp    $0x20,%ebp
  je     L20
L36:
  movsbl (%rbx),%edx
  xor    %eax,%eax
  mov    %ebp,%esi
  mov    %r12,%rdi
  call   <printf@plt>
  cmpb   $0x0,(%rbx)
  jne    L13
  lea    D25(%rip),%rdi
  call   <puts@plt>
L20:
  xor    %ebx,%ebx
  lea    0x6(%rsp),%rbp
  lea    D7(%rip),%r12
  nopw   0x0(%rax,%rax,1)
L37:
  movsbl 0x0(%rbp,%rbx,1),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xa,%rbx
  jne    L37
  lea    D18(%rip),%r13
  mov    %rbp,%rsi
  xor    %eax,%eax
  xor    %ebx,%ebx
  mov    %r13,%rdi
  call   <printf@plt>
  mov    %rbp,%rsi
  mov    %r13,%rdi
  xor    %eax,%eax
  movb   $0x0,0xa(%rsp)
  call   <printf@plt>
  nopl   0x0(%rax,%rax,1)
L15:
  movsbl 0x0(%rbp,%rbx,1),%edx
  mov    %ebx,%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0xa,%rbx
  jne    L15
  lea    D70(%rip),%r12
  lea    D76(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  mov    %r12,%rsi
  lea    D31(%rip),%rbp
  call   <printf@plt>
  mov    $0x73,%edx
L8:
  mov    %ebx,%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    $0x5,%rbx
  je     L7
  movsbl (%r12,%rbx,1),%edx
  jmp    L8
  xchg   %ax,%ax
L7:
  mov    0x38(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L17
  add    $0x48,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
L17:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D84
datatype: string
name: retrieve_data
arguments: label: D58
datatype: string
name: retrieve_data
arguments: label: D15
datatype: string
name: retrieve_data
arguments: label: D81
datatype: string
name: retrieve_data
arguments: label: D25
datatype: string
name: retrieve_data
arguments: label: D7
datatype: string
name: retrieve_data
arguments: label: D18
datatype: string
name: retrieve_data
arguments: label: D70
datatype: string
name: retrieve_data
arguments: label: D76
datatype: string
name: retrieve_data
arguments: label: D31
datatype: string

tool: D84: str1: 

tool: D58: str1[%d]: '%c'


tool: D15: %s

tool: D81: str1 = "%s"


tool: D25: str1 end.

tool: D7: str2[%d]: '%c'


tool: D18: str2 = "%s"


tool: D70: str3

tool: D76: str3 = "%s"


tool: D31: *(str3 + %d): '%c'


assistant: ```c
int main(void) {
  int i;
  char str1[32];
  char str2[] = "str2ABCDE";
  char *str3 = "str3";
  printf("str1: ");
  scanf("%s", str1);
  printf("str1 = \"%s\"\n", str1);
  for (i = 0; i < 32; i++) {
    printf("str1[%d]: \'%c\'\n", i, str1[i]);
    if (str1[i] == '\0') {
      printf("str1 end.\n");
      break;
    }
  }
  for (i = 0; i < 10; i++) {
    printf("str2[%d]: \'%c\'\n", i, str2[i]);
  }
  printf("str2 = \"%s\"\n", str2);
  str2[4] = '\0';
  printf("str2 = \"%s\"\n", str2);
  for (i = 0; i < 10; i++) {
    printf("str2[%d]: \'%c\'\n", i, str2[i]);
  }
  printf("str3 = \"%s\"\n", str3);
  for (i = 0; i < 5; i++) {
    printf("*(str3 + %d): \'%c\'\n", i, *(str3 + i));
  }
  return 0;
}
```