user: 这段汇编代码用C语言表示是什么样的？

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x18,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x8(%rsp)
  xor    %eax,%eax
  movl   $0x1,0x4(%rsp)
  cmp    $0x1,%edi
  jle    L1
  lea    0x8(%rsi),%rbx
  lea    -0x2(%rdi),%eax
  lea    0x10(%rsi,%rax,8),%r15
  mov    $0x0,%r13d
  mov    $0x0,%r12d
  mov    $0x0,%r14d
  mov    $0x1,%ebp
L71:
  mov    $0x6d,%esi
  mov    (%rbx),%rdi
  call   <Strchr@plt>
  test   %rax,%rax
  cmovne %ebp,%r14d
  mov    $0x64,%esi
  mov    (%rbx),%rdi
  call   <Strchr@plt>
  test   %rax,%rax
  cmovne %ebp,%r12d
  mov    $0x68,%esi
  mov    (%rbx),%rdi
  call   <Strchr@plt>
  test   %rax,%rax
  cmovne %ebp,%r13d
  add    $0x8,%rbx
  cmp    %r15,%rbx
  jne    L71
  test   %r13d,%r13d
  jne    L62
  test   %r14d,%r14d
  je     L27
  lea    D58(%rip),%rdi
  call   <puts@plt>
L24:
  test   %r12d,%r12d
  jne    L89
L23:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L28
  mov    $0x0,%eax
  add    $0x18,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L62:
  lea    D40(%rip),%rdi
  call   <puts@plt>
  jmp    L23
L1:
  mov    $0x0,%r12d
L27:
  lea    D57(%rip),%rdi
  call   <puts@plt>
  jmp    L24
L89:
  mov    $0x4,%esi
  lea    D2(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    D44(%rip),%rdi
  call   <puts@plt>
  lea    D63(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x4(%rsp),%rbx
  lea    0x8(%rsp),%r12
  lea    D20(%rip),%rbp
L44:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%rbx
  cmp    %rbx,%r12
  jne    L44
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L23
L28:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D58
datatype: string
name: read_data
arguments: data_tag: D40
datatype: string
name: read_data
arguments: data_tag: D57
datatype: string
name: read_data
arguments: data_tag: D2
datatype: string
name: read_data
arguments: data_tag: D44
datatype: string
name: read_data
arguments: data_tag: D63
datatype: string
name: read_data
arguments: data_tag: D20
datatype: string

tool: D58: __LITTLE_ENDIAN_BITFIELD

tool: D40: -m	macro output
-d	debug
-h	help

tool: D57: little endian

tool: D2: sizeof(unsigned int) = %d


tool: D44: unsigned int test = 1;

tool: D63: in memory as: 

tool: D20: %02x 

assistant: ```c
int main(int argc, char **argv) {
  unsigned int test = 1;
  unsigned char *x;
  int macro = 0, debug = 0, help = 0, j;
  for (j = 1; j < argc; j++) {
    if (Strchr(argv[j], 'm'))
      macro = 1;
    if (Strchr(argv[j], 'd'))
      debug = 1;
    if (Strchr(argv[j], 'h'))
      help = 1;
  }
  if (help) {
    printf("-m	macro output\n"
           "-d	debug\n"
           "-h	help\n");
    return 0;
  }
  x = (unsigned char *)&test;
  if (*x == 0x00) {
    if (macro)
      printf("__BIG_ENDIAN_BITFIELD\n");
    else
      printf("big endian\n");
  } else if (*x == 0x01) {
    if (macro)
      printf("__LITTLE_ENDIAN_BITFIELD\n");
    else
      printf("little endian\n");
  } else {
    printf("\nWARNING!!! byteorder exception\n\n");
    debug = 1;
  }
  if (debug) {
    printf("sizeof(unsigned int) = %d\n", (int)sizeof(unsigned int));
    printf("unsigned int test = 1;\n");
    printf("in memory as: ");
    for (j = 0; j < sizeof(unsigned int); j++)
      printf("%02x ", x[j]);
    printf("\n");
  }
  return 0;
}
```