user: 下面的汇编指令对应的C源代码是什么：

```asm
<parse_four>:
  endbr64
  push   %r15
  mov    %r8,%r15
  push   %r14
  mov    %rcx,%r14
  push   %r13
  push   %r12
  push   %rbp
  lea    D0(%rip),%rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,0x10(%rsp)
  mov    %rbp,%rsi
  mov    %rdx,0x18(%rsp)
  call   <strtok@plt>
  test   %rax,%rax
  je     L89
  mov    %rax,%r13
  xor    %ebx,%ebx
  xor    %r12d,%r12d
  jmp    L95
  nopl   0x0(%rax)
L91:
  cmp    $0x2,%r12d
  jne    L5
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    $0x7ff,%edx
  mov    %r13,%rsi
  mov    0x18(%rsp),%r13
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    %r13,%rdi
  add    $0x2,%ebx
  movslq %eax,%rdx
  mov    %rdx,0x8(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%rdx
  movb   $0x0,0x0(%r13,%rdx,1)
  nopw   0x0(%rax,%rax,1)
L1:
  mov    %rbp,%rsi
  xor    %edi,%edi
  call   <strtok@plt>
  mov    %rax,%r13
  test   %rax,%rax
  je     L111
L95:
  add    $0x1,%r12d
  cmp    $0x3,%r12d
  je     L87
  jg     L69
  cmp    $0x1,%r12d
  jne    L91
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    $0x7ff,%edx
  mov    %r13,%rsi
  mov    0x10(%rsp),%r13
  cmp    %edx,%eax
  cmovg  %edx,%eax
  mov    %r13,%rdi
  add    $0x1,%ebx
  movslq %eax,%rdx
  mov    %rdx,0x8(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%rdx
  movb   $0x0,0x0(%r13,%rdx,1)
  jmp    L1
  cs nopw 0x0(%rax,%rax,1)
L69:
  cmp    $0x4,%r12d
  jne    L5
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    $0x7ff,%edx
  mov    %r13,%rsi
  mov    %r15,%rdi
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    $0x8,%ebx
  movslq %eax,%rdx
  mov    %rdx,0x8(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%rdx
  movb   $0x0,(%r15,%rdx,1)
  jmp    L1
  nopl   (%rax)
L5:
  xor    %ebx,%ebx
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L87:
  mov    %r13,%rdi
  call   <strlen@plt>
  mov    $0x7ff,%edx
  mov    %r13,%rsi
  mov    %r14,%rdi
  cmp    %edx,%eax
  cmovg  %edx,%eax
  add    $0x4,%ebx
  movslq %eax,%rdx
  mov    %rdx,0x8(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%rdx
  movb   $0x0,(%r14,%rdx,1)
  jmp    L1
  nopw   0x0(%rax,%rax,1)
L111:
  xor    %eax,%eax
  cmp    $0xf,%ebx
  setne  %al
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L89:
  add    $0x28,%rsp
  mov    $0x1,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

tool_calls: name: parse_data
arguments: tag: D0
type_infered: string

tool: D0: 	


assistant: ```c
int parse_four(char *line, char *field1, char *field2, char *field3, char *field4) {
  char *ptr;
  int i, n, rc;
  i = 0;
  rc = 0;
  ptr = strtok(line, "\t\n");
  while (ptr != NULL) {
    i++;
    switch (i) {
    case 1:
      n = (int)strlen(ptr);
      if (n > 2048 - 1)
        n = 2048 - 1;
      strncpy(field1, ptr, (size_t)n);
      field1[n] = (char)'\0';
      rc += 1;
      break;
    case 2:
      n = (int)strlen(ptr);
      if (n > 2048 - 1)
        n = 2048 - 1;
      strncpy(field2, ptr, (size_t)n);
      field2[n] = (char)'\0';
      rc += 2;
      break;
    case 3:
      n = (int)strlen(ptr);
      if (n > 2048 - 1)
        n = 2048 - 1;
      strncpy(field3, ptr, (size_t)n);
      field3[n] = (char)'\0';
      rc += 4;
      break;
    case 4:
      n = (int)strlen(ptr);
      if (n > 2048 - 1)
        n = 2048 - 1;
      strncpy(field4, ptr, (size_t)n);
      field4[n] = (char)'\0';
      rc += 8;
      break;
    default:
      rc = 0;
    }
    ptr = strtok(NULL, "\t\n");
  }
  if (rc == 15)
    return (0);
  else
    return (1);
}
```