user: 请从给定的汇编指令推导出C源代码：

```asm
<getop>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x18,%rsp
  nopl   (%rax)
L272:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %al,(%rbx)
  mov    %eax,%r12d
  mov    %eax,%edi
  cmp    $0x9,%eax
  je     L272
  cmp    $0x20,%al
  je     L272
  movb   $0x0,0x1(%rbx)
  mov    %eax,%r14d
  cmp    $0x2f,%eax
  ja     L144
  movabs $0xffff77dffffffbff,%rax
  bt     %r12,%rax
  jb     L253
L71:
  add    $0x18,%rsp
  mov    %r12d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   (%rax)
L144:
  mov    D0(%rip),%r13
L51:
  movsbl %dil,%ebp
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L134
  sub    $0x2d,%r14d
  cmp    $0x1,%r14d
  jbe    L134
  add    $0x1,%rbx
  jmp    L258
  nop
L48:
  cmp    %eax,0x0(%r13)
  je     L97
  lea    -0x9(%rax),%edx
  add    $0x1,%rbx
  cmp    $0x1,%edx
  jbe    L97
L258:
  xor    %eax,%eax
  call   <getch@plt>
  mov    %rbx,%rcx
  mov    %al,(%rbx)
  cmp    $0x20,%al
  jne    L48
L97:
  movb   $0x0,(%rcx)
  mov    $0x31,%r12d
  cmp    %eax,0x0(%r13)
  je     L71
  mov    %eax,%edi
  call   <ungetch@plt>
  jmp    L71
L253:
  cmp    $0x2a,%r12d
  je     L71
  cmp    $0x2d,%r12d
  jne    L144
  xor    %eax,%eax
  call   <getch@plt>
  mov    %eax,%r14d
  lea    -0x9(%rax),%eax
  cmp    $0x1,%eax
  jbe    L246
  cmp    $0x20,%r14d
  je     L246
  mov    D0(%rip),%r13
  cmp    %r14d,0x0(%r13)
  je     L246
  mov    %r14d,%edi
  mov    %r14d,%r12d
  call   <ungetch@plt>
  mov    %r14d,%edi
  jmp    L51
  xchg   %ax,%ax
L134:
  mov    %ebp,%edi
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L188
  mov    %rbx,%rbp
  xor    %r14d,%r14d
L76:
  cmp    $0x2e,%r12d
  je     L46
L12:
  movb   $0x0,0x0(%rbp)
  cmp    %r12d,0x0(%r13)
  je     L26
  mov    %r12d,%edi
  mov    $0x30,%r12d
  call   <ungetch@plt>
  jmp    L71
L188:
  mov    %ebx,%eax
  lea    0x1(%rbx),%r15
  neg    %eax
  mov    %eax,0xc(%rsp)
  nopl   0x0(%rax)
L238:
  xor    %eax,%eax
  mov    %r15,%rbp
  call   <getch@plt>
  mov    %eax,%r12d
  mov    0xc(%rsp),%eax
  mov    %r12b,(%r15)
  movsbl %r12b,%edi
  lea    (%rax,%r15,1),%r14d
  add    $0x1,%r15
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L238
  jmp    L76
L26:
  mov    $0x30,%r12d
  jmp    L71
L46:
  movslq %r14d,%r14
  lea    0x1(%rbx,%r14,1),%rbx
  nopw   0x0(%rax,%rax,1)
L4:
  xor    %eax,%eax
  mov    %rbx,%rbp
  add    $0x1,%rbx
  call   <getch@plt>
  mov    %al,-0x1(%rbx)
  movsbl %al,%edi
  mov    %eax,%r12d
  call   <isdigit@plt>
  test   %rax,%rax
  jne    L4
  jmp    L12
L246:
  mov    %r14d,%edi
  call   <ungetch@plt>
  jmp    L71
```

assistant: ```c
int getop(char s[]) {
  int i, c;
  char word[10];
  while ((s[0] = c = getch()) == ' ' || c == '\t')
    ;
  s[1] = '\0';
  if (c == '%' || c == '+' || c == '/' || c == '\n' || c == '*') {
    return c;
  }
  i = 0;
  if (c == '-') {
    if ((c = getch()) == ' ' || c == '\t' || c == EOF || c == '\n') {
      ungetch(c);
      return '-';
    } else
      ungetch(c);
  }
  if (!isdigit(c) && c != '.' && c != '-') {
    while ((s[++i] = c = getch()) != ' ' && c != EOF && c != '\n' && c != '\t')
      ;
    s[i] = '\0';
    if (c != EOF)
      ungetch(c);
    return '1';
  }
  if (isdigit(c))
    while (isdigit(s[++i] = c = getch()))
      ;
  if (c == '.') {
    while (isdigit(s[++i] = c = getch()))
      ;
  }
  s[i] = '\0';
  if (c != EOF)
    ungetch(c);
  return '0';
}
```