user: 提供的汇编代码用C语言表示会是什么样子:

```asm
<itoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x60,%rsp
  mov    %edi,-0x54(%rbp)
  mov    %rsi,-0x60(%rbp)
  mov    %edx,-0x58(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movabs $0x3736353433323130,%rax
  movabs $0x4645444342413938,%rdx
  mov    %rax,-0x30(%rbp)
  mov    %rdx,-0x28(%rbp)
  movabs $0x4e4d4c4b4a494847,%rax
  movabs $0x565554535251504f,%rdx
  mov    %rax,-0x20(%rbp)
  mov    %rdx,-0x18(%rbp)
  movl   $0x5a595857,-0x10(%rbp)
  movb   $0x0,-0xc(%rbp)
  movl   $0x0,-0x3c(%rbp)
  cmpl   $0xa,-0x58(%rbp)
  jne    L0
  cmpl   $0x0,-0x54(%rbp)
  jns    L0
  mov    -0x54(%rbp),%eax
  neg    %eax
  mov    %eax,-0x40(%rbp)
  mov    -0x3c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2d,(%rax)
  jmp    L11
L0:
  mov    -0x54(%rbp),%eax
  mov    %eax,-0x40(%rbp)
L11:
  mov    -0x58(%rbp),%ecx
  mov    -0x40(%rbp),%eax
  mov    $0x0,%edx
  div    %ecx
  mov    %edx,%ecx
  mov    -0x3c(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x3c(%rbp)
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rax,%rdx
  mov    %ecx,%eax
  movzbl -0x30(%rbp,%rax,1),%eax
  mov    %al,(%rdx)
  mov    -0x58(%rbp),%esi
  mov    -0x40(%rbp),%eax
  mov    $0x0,%edx
  div    %esi
  mov    %eax,-0x40(%rbp)
  cmpl   $0x0,-0x40(%rbp)
  jne    L11
  mov    -0x3c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x60(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x2d,%al
  jne    L39
  movl   $0x1,-0x34(%rbp)
  jmp    L2
L39:
  movl   $0x0,-0x34(%rbp)
L2:
  mov    -0x34(%rbp),%eax
  mov    %eax,-0x38(%rbp)
  jmp    L38
L63:
  mov    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0x41(%rbp)
  mov    -0x3c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  sub    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rdx,%rax
  mov    -0x38(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x60(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x3c(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    -0x34(%rbp),%eax
  add    %edx,%eax
  sub    -0x38(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x60(%rbp),%rax
  add    %rax,%rdx
  movzbl -0x41(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x38(%rbp)
L38:
  mov    -0x3c(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  cmp    %eax,-0x38(%rbp)
  jle    L63
  mov    -0x60(%rbp),%rax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L86
  call   <__stack_chk_fail@plt>
L86:
  leave
  ret
```

assistant: ```c
char *itoa(int num, char *str, int radix) {
  char index[] = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ";
  unsigned unum;
  int i = 0, j, k;
  if (radix == 10 && num < 0) {
    unum = (unsigned)-num;
    str[i++] = '-';
  } else
    unum = (unsigned)num;
  do {
    str[i++] = index[unum % (unsigned)radix];
    unum /= radix;
  } while (unum);
  str[i] = '\0';
  if (str[0] == '-')
    k = 1;
  else
    k = 0;
  char temp;
  for (j = k; j <= (i - 1) / 2; j++) {
    temp = str[j];
    str[j] = str[i - 1 + k - j];
    str[i - 1 + k - j] = temp;
  }
  return str;
}
```