user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  lea    D8(%rip),%r15
  push   %r14
  push   %r13
  push   %r12
  lea    D5(%rip),%r12
  push   %rbp
  push   %rbx
  sub    $0x418,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x408(%rsp)
  xor    %eax,%eax
  mov    %rsp,%rbp
L56:
  lea    D37(%rip),%r14
L130:
  mov    $0x400,%esi
  mov    %rbp,%rdi
  call   <getl@plt>
  mov    %eax,%r13d
  test   %eax,%eax
  jle    L137
  mov    %rbp,%rsi
  mov    %r15,%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L86
  nopl   0x0(%rax,%rax,1)
L6:
  movsbl 0x0(%rbp,%rbx,1),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
L86:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L6
  mov    $0xa,%edi
  call   <putchar@plt>
  lea    -0x1(%r13),%eax
  cltq
  cmpb   $0xa,(%rsp,%rax,1)
  je     L55
L21:
  cmp    $0x2,%r13d
  jle    L130
  mov    %rbp,%rsi
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  xor    %ebx,%ebx
  call   <printf@plt>
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L13
  nopw   0x0(%rax,%rax,1)
L18:
  movsbl 0x0(%rbp,%rbx,1),%esi
  mov    %r12,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
L13:
  mov    %rbp,%rdi
  call   <strlen@plt>
  cmp    %ebx,%eax
  jg     L18
  lea    D24(%rip),%rdi
  call   <puts@plt>
  jmp    L56
  nopl   0x0(%rax,%rax,1)
L55:
  lea    -0x2(%r13),%eax
  cmp    $0x1,%r13d
  je     L130
  cltq
  jmp    L10
  nopw   0x0(%rax,%rax,1)
L131:
  sub    $0x1,%rax
  cmp    $0xffffffff,%eax
  je     L21
L10:
  movzbl 0x0(%rbp,%rax,1),%edx
  mov    %eax,%ecx
  cmp    $0x20,%dl
  je     L131
  cmp    $0x9,%dl
  je     L131
  test   %eax,%eax
  jle    L21
  lea    0x1(%rax),%eax
  cltq
  movb   $0xa,(%rsp,%rax,1)
  lea    0x2(%rcx),%eax
  cltq
  movb   $0x0,(%rsp,%rax,1)
  jmp    L21
L137:
  mov    0x408(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L107
  add    $0x418,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L107:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: retrieve_data
arguments: label: D8
type_infered: string
name: retrieve_data
arguments: label: D5
type_infered: string
name: retrieve_data
arguments: label: D37
type_infered: string
name: retrieve_data
arguments: label: D12
type_infered: string
name: retrieve_data
arguments: label: D24
type_infered: string

tool: D8: Before remove trailing blanks and tabs: %s

tool: D5:  %d 

tool: D37: Charaters in char array: 

tool: D12: After remove trailing blanks and tabs: %s

tool: D24: 


assistant: ```c
int main() {
  int len;
  char line[1024];
  while ((len = getl(line, 1024)) > 0) {
    printf("Before remove trailing blanks and tabs: %s", line);
    printf("Charaters in char array: ");
    int j;
    for (j = 0; j < strlen(line); ++j) {
      printf(" %d ", line[j]);
    }
    printf("\n");
    if (line[len - 1] == '\n') {
      int i = len - 2;
      while (i >= 0 && (line[i] == ' ' || line[i] == '\t')) {
        --i;
      }
      if (i > 0) {
        line[i + 1] = '\n';
        line[i + 2] = '\0';
      }
    }
    if (len > 2) {
      printf("After remove trailing blanks and tabs: %s", line);
      printf("Charaters in char array: ");
      for (j = 0; j < strlen(line); ++j) {
        printf(" %d ", line[j]);
      }
      printf("\n\n");
    }
  }
}
```