user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x40,%rsp
  mov    %edi,-0x1034(%rbp)
  mov    %rsi,-0x1040(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x1024(%rbp)
  movl   $0x0,-0x1020(%rbp)
  jmp    L151
L167:
  mov    -0x1020(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x1040(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  je     L0
  mov    -0x1020(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x1040(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L82
L0:
  movl   $0x0,-0x1024(%rbp)
L82:
  addl   $0x1,-0x1020(%rbp)
L151:
  mov    -0x1020(%rbp),%eax
  cmp    -0x1034(%rbp),%eax
  jl     L167
  cmpl   $0x0,-0x1024(%rbp)
  je     L150
  mov    D5(%rip),%rax
  mov    (%rax),%edx
  mov    -0x1040(%rbp),%rax
  add    $0x8,%rax
  mov    (%rax),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <open@plt>
  mov    %eax,-0x1014(%rbp)
  jmp    L3
L55:
  mov    -0x1018(%rbp),%eax
  mov    %eax,-0x1028(%rbp)
L3:
  lea    -0x1010(%rbp),%rcx
  mov    -0x1014(%rbp),%eax
  mov    $0x1000,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x1018(%rbp)
  cmpl   $0x0,-0x1018(%rbp)
  jg     L55
  mov    -0x1014(%rbp),%eax
  mov    %eax,%edi
  call   <close@plt>
  jmp    L72
L74:
  mov    -0x1018(%rbp),%eax
  mov    %eax,-0x1028(%rbp)
L150:
  mov    D17(%rip),%rax
  mov    (%rax),%eax
  lea    -0x1010(%rbp),%rcx
  mov    $0x1000,%edx
  mov    %rcx,%rsi
  mov    %eax,%edi
  call   <read@plt>
  mov    %eax,-0x1018(%rbp)
  cmpl   $0x0,-0x1018(%rbp)
  jg     L74
  jmp    L72
L138:
  movl   $0x0,-0x101c(%rbp)
  jmp    L28
L14:
  mov    -0x101c(%rbp),%eax
  cltq
  movzbl -0x1010(%rbp,%rax,1),%eax
  test   %al,%al
  je     L139
  mov    -0x101c(%rbp),%eax
  cltq
  movzbl -0x1010(%rbp,%rax,1),%eax
  cmp    $0xa,%al
  jne    L63
L139:
  mov    -0x101c(%rbp),%eax
  cltq
  movb   $0x30,-0x1010(%rbp,%rax,1)
L63:
  addl   $0x1,-0x101c(%rbp)
L28:
  mov    -0x101c(%rbp),%eax
  cmp    -0x1028(%rbp),%eax
  jle    L14
  addl   $0x1,-0x1028(%rbp)
L72:
  mov    -0x1028(%rbp),%eax
  and    $0x7,%eax
  test   %eax,%eax
  jne    L138
  mov    $0x0,%edi
  call   <printHandler@plt>
  jmp    L157
L101:
  lea    -0x1010(%rbp),%rax
  mov    %rax,%rdi
  call   <conversionManager@plt>
  subl   $0x8,-0x1028(%rbp)
L157:
  cmpl   $0x0,-0x1028(%rbp)
  jg     L101
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L6
  call   <__stack_chk_fail@plt>
L6:
  leave
  ret
```

assistant: ```c
int main(int argc, char *argv[]) {
  char buffer[4096];
  int charCount;
  int n;
  int status = 1;
  int i;
  for (i = 0; i < argc; i++) {
    if (*argv[i] == '-' || *argv[i] == '\0')
      status = 0;
  }
  if (status) {
    int file_descriptor = open(argv[1], O_RDONLY);
    while ((n = read(file_descriptor, buffer, 4096)) > 0) {
      charCount = n;
    }
    close(file_descriptor);
  } else {
    while ((n = read(STDIN_FILENO, buffer, 4096)) > 0) {
      charCount = n;
    }
  }
  while ((charCount % 8) != 0) {
    int i;
    for (i = 0; i <= charCount; i++) {
      if (buffer[i] == '\0' || buffer[i] == '\n')
        buffer[i] = '0';
    }
    charCount++;
  }
  printHandler(0);
  while (charCount >= 1) {
    conversionManager(buffer);
    charCount = charCount - 8;
  }
  return 0;
}
```