user: 这段汇编代码用C语言表示是什么样的?

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x8000(%rsp),%r11
L46:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L46
  sub    $0x30,%rsp
  mov    %edi,-0x8024(%rbp)
  mov    %rsi,-0x8030(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x65686361636c63,%rax
  mov    $0x0,%edx
  mov    %rax,-0x8010(%rbp)
  mov    %rdx,-0x8008(%rbp)
  lea    -0x8000(%rbp),%rax
  mov    $0x7ff0,%edx
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  movl   $0x6,-0x801c(%rbp)
  movl   $0x2,-0x8018(%rbp)
  jmp    L13
L77:
  addl   $0x1,-0x801c(%rbp)
  mov    -0x801c(%rbp),%eax
  cltq
  movb   $0x20,-0x8010(%rbp,%rax,1)
  movl   $0x0,-0x8014(%rbp)
  jmp    L56
L83:
  mov    -0x8018(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8030(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8014(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5c,%al
  jne    L14
  addl   $0x1,-0x801c(%rbp)
  mov    -0x801c(%rbp),%eax
  cltq
  movb   $0x5c,-0x8010(%rbp,%rax,1)
  addl   $0x1,-0x801c(%rbp)
  mov    -0x801c(%rbp),%eax
  cltq
  movb   $0x5c,-0x8010(%rbp,%rax,1)
  jmp    L95
L14:
  mov    -0x8018(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8030(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8014(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L6
  addl   $0x1,-0x801c(%rbp)
  mov    -0x801c(%rbp),%eax
  cltq
  movb   $0x5c,-0x8010(%rbp,%rax,1)
  addl   $0x1,-0x801c(%rbp)
  mov    -0x801c(%rbp),%eax
  cltq
  movb   $0x22,-0x8010(%rbp,%rax,1)
  jmp    L95
L6:
  mov    -0x8018(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8030(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8014(%rbp),%eax
  cltq
  add    %rdx,%rax
  addl   $0x1,-0x801c(%rbp)
  movzbl (%rax),%edx
  mov    -0x801c(%rbp),%eax
  cltq
  mov    %dl,-0x8010(%rbp,%rax,1)
L95:
  addl   $0x1,-0x8014(%rbp)
L56:
  mov    -0x8018(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x8030(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x8014(%rbp),%eax
  cltq
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L83
  addl   $0x1,-0x8018(%rbp)
L13:
  mov    -0x8018(%rbp),%eax
  cmp    -0x8024(%rbp),%eax
  jl     L77
  addl   $0x1,-0x801c(%rbp)
  mov    -0x801c(%rbp),%eax
  cltq
  movb   $0x0,-0x8010(%rbp,%rax,1)
  lea    -0x8010(%rbp),%rax
  mov    %rax,%rdi
  call   <system@plt>
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L4
  call   <__stack_chk_fail@plt>
L4:
  leave
  ret
```

assistant: ```c
int main(int pArgC, char *pArgV[]) {
  char clcacheCommand[32768] = "clcache";
  int k = 6;
  for (int i = 2; i < pArgC; ++i) {
    clcacheCommand[++k] = ' ';
    for (int j = 0; pArgV[i][j]; ++j) {
      if (pArgV[i][j] == '\\') {
        clcacheCommand[++k] = '\\';
        clcacheCommand[++k] = '\\';
      } else if (pArgV[i][j] == '"') {
        clcacheCommand[++k] = '\\';
        clcacheCommand[++k] = '"';
      } else {
        clcacheCommand[++k] = pArgV[i][j];
      }
    }
  }
  clcacheCommand[++k] = '\0';
  return system(clcacheCommand);
}
```