user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<main>:
  endbr64
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebx
  mov    D6(%rip),%rax
  cmp    (%rax),%bl
  je     L142
  mov    $0x0,%r12d
  mov    $0x0,%r13d
  jmp    L105
L337:
  cmp    $0x22,%bl
  je     L179
  cmp    $0x27,%bl
  je     L214
  movsbl %bl,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
L233:
  mov    D6(%rip),%rax
  cmp    %bl,(%rax)
  je     L142
L105:
  mov    $0x0,%eax
  call   <getchar@plt>
  mov    %eax,%ebp
  test   %r12d,%r12d
  jne    L341
  cmp    $0x2f,%bl
  sete   %al
  cmp    $0x2a,%bpl
  jne    L67
  test   %al,%al
  jne    L261
L67:
  cmp    $0x2f,%bpl
  jne    L337
  test   %al,%al
  je     L337
  mov    $0x2f,%ebx
  mov    $0x3,%r12d
  jmp    L233
L179:
  mov    $0x22,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  mov    $0x5,%r12d
  jmp    L233
L214:
  mov    $0x27,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  mov    $0x4,%r12d
  jmp    L233
L341:
  cmp    $0x5,%r12d
  je     L270
  cmp    $0x4,%r12d
  je     L2
  cmp    $0x2a,%bl
  sete   %dl
  cmp    $0x1,%r12d
  sete   %al
  test   %al,%dl
  je     L295
  cmp    $0x2f,%bpl
  je     L80
L295:
  cmp    $0x2,%r12d
  je     L6
  cmp    $0xa,%bl
  jne    L189
  cmp    $0x3,%r12d
  je     L176
L189:
  cmp    $0x6,%r12d
  je     L351
  cmp    $0x7,%r12d
  je     L19
  mov    %ebp,%ebx
  jmp    L233
L270:
  movsbl %bl,%edi
  call   <putchar@plt>
  cmp    $0x5c,%bl
  je     L212
  cmp    $0x22,%bl
  je     L17
  mov    %ebp,%ebx
  jmp    L233
L17:
  mov    %ebp,%ebx
  mov    %r13d,%r12d
  jmp    L233
L2:
  movsbl %bl,%edi
  call   <putchar@plt>
  cmp    $0x5c,%bl
  je     L215
  cmp    $0x27,%bl
  setne  %r12b
  movzbl %r12b,%r12d
  shl    $0x2,%r12d
  mov    %ebp,%ebx
  jmp    L233
L351:
  movsbl %bl,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  mov    $0x5,%r12d
  jmp    L233
L19:
  movsbl %bl,%edi
  call   <putchar@plt>
  mov    %ebp,%ebx
  mov    $0x4,%r12d
  jmp    L233
L261:
  mov    $0x2a,%ebx
  mov    $0x1,%r12d
  jmp    L233
L212:
  mov    %ebp,%ebx
  mov    $0x6,%r12d
  jmp    L233
L215:
  mov    %ebp,%ebx
  mov    $0x7,%r12d
  jmp    L233
L80:
  mov    $0x2f,%ebx
  mov    $0x2,%r12d
  jmp    L233
L6:
  mov    %ebp,%ebx
  mov    %r13d,%r12d
  jmp    L233
L176:
  mov    %ebp,%ebx
  mov    %r13d,%r12d
  jmp    L233
L142:
  mov    $0x0,%eax
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
int main() {
  char buf[2];
  int state;
  state = 0;
  buf[1] = getchar();
  while ((buf[0] = buf[1]) != EOF) {
    buf[1] = getchar();
    if (state == 0) {
      if (buf[0] == '/' && buf[1] == '*') {
        state = 1;
      } else if (buf[0] == '/' && buf[1] == '/') {
        state = 3;
      } else if (buf[0] == '"') {
        state = 5;
        putchar(buf[0]);
      } else if (buf[0] == '\'') {
        state = 4;
        putchar(buf[0]);
      } else {
        putchar(buf[0]);
      }
    } else if (state == 5) {
      putchar(buf[0]);
      if (buf[0] == '\\')
        state = 6;
      else if (buf[0] == '"')
        state = 0;
    } else if (state == 4) {
      putchar(buf[0]);
      if (buf[0] == '\\')
        state = 7;
      else if (buf[0] == '\'')
        state = 0;
    } else if (state == 1 && buf[0] == '*' && buf[1] == '/') {
      state = 2;
    } else if (state == 2) {
      state = 0;
    } else if (state == 3 && buf[0] == '\n') {
      state = 0;
    } else if (state == 6) {
      state = 5;
      putchar(buf[0]);
    } else if (state == 7) {
      state = 4;
      putchar(buf[0]);
    }
  }
}
```