user: 请从给定的汇编指令推导出C源代码：

```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    L93
  lea    -0x2(%rdi),%eax
  lea    0x8(%rsi),%rbx
  xor    %r12d,%r12d
  xor    %r13d,%r13d
  lea    0x10(%rsi,%rax,8),%r15
  xor    %r14d,%r14d
  mov    $0x1,%ebp
  nopl   (%rax)
L45:
  mov    (%rbx),%rdi
  mov    $0x6d,%esi
  call   <Strchr@plt>
  mov    (%rbx),%rdi
  mov    $0x64,%esi
  test   %rax,%rax
  cmovne %ebp,%r14d
  call   <Strchr@plt>
  mov    (%rbx),%rdi
  mov    $0x68,%esi
  test   %rax,%rax
  cmovne %ebp,%r13d
  call   <Strchr@plt>
  test   %rax,%rax
  cmovne %ebp,%r12d
  add    $0x8,%rbx
  cmp    %r15,%rbx
  jne    L45
  test   %r12d,%r12d
  jne    L26
  test   %r14d,%r14d
  je     L105
  lea    D67(%rip),%rdi
  call   <puts@plt>
L94:
  test   %r13d,%r13d
  jne    L28
L32:
  mov    0x8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L15
  add    $0x18,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L105:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  jmp    L94
L28:
  mov    $0x4,%esi
  lea    D76(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    D18(%rip),%rdi
  lea    0x4(%rsp),%rbx
  call   <puts@plt>
  lea    D78(%rip),%rdi
  xor    %eax,%eax
  lea    0x8(%rsp),%r12
  call   <printf@plt>
  lea    D10(%rip),%rbp
L61:
  movzbl (%rbx),%esi
  mov    %rbp,%rdi
  xor    %eax,%eax
  add    $0x1,%rbx
  call   <printf@plt>
  cmp    %r12,%rbx
  jne    L61
  mov    $0xa,%edi
  call   <putchar@plt>
  jmp    L32
L26:
  lea    D25(%rip),%rdi
  call   <puts@plt>
  jmp    L32
L93:
  lea    D15(%rip),%rdi
  call   <puts@plt>
  jmp    L32
L15:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: label: D67
type_guessed: string
name: parse_data
arguments: label: D15
type_guessed: string
name: parse_data
arguments: label: D76
type_guessed: string
name: parse_data
arguments: label: D18
type_guessed: string
name: parse_data
arguments: label: D78
type_guessed: string
name: parse_data
arguments: label: D10
type_guessed: string
name: parse_data
arguments: label: D25
type_guessed: string

tool: D67: __LITTLE_ENDIAN_BITFIELD

tool: D15: little endian

tool: D76: sizeof(unsigned int) = %d


tool: D18: unsigned int test = 1;

tool: D78: in memory as: 

tool: D10: %02x 

tool: D25: -m	macro output
-d	debug
-h	help

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", 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;
}
```