user: 下面的汇编指令对应的C源代码是什么:

```asm
<longestPalindrome>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x14(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x14(%rbp),%eax
  add    $0x2,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x8(%rbp)
  mov    -0x10(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rax,%rsi
  mov    %rdx,%rdi
  call   <strcpy@plt>
  mov    -0x10(%rbp),%rax
  movb   $0xff,(%rax)
  movl   $0x1,-0x28(%rbp)
  movl   $0x1,-0x20(%rbp)
  mov    -0x20(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  movl   $0x1,-0x2c(%rbp)
  jmp    L38
L73:
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  mov    -0x2c(%rbp),%eax
  mov    %eax,-0x18(%rbp)
  jmp    L1
L85:
  addl   $0x1,-0x18(%rbp)
L1:
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L85
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  jmp    L46
L39:
  subl   $0x1,-0x1c(%rbp)
  addl   $0x1,-0x18(%rbp)
L46:
  mov    -0x1c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x18(%rbp),%eax
  cltq
  lea    0x1(%rax),%rcx
  mov    -0x10(%rbp),%rax
  add    %rcx,%rax
  movzbl (%rax),%eax
  cmp    %al,%dl
  je     L39
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  cmp    %eax,-0x28(%rbp)
  jg     L71
  mov    -0x18(%rbp),%eax
  sub    -0x1c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x28(%rbp)
  mov    -0x1c(%rbp),%eax
  mov    %eax,-0x24(%rbp)
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x20(%rbp)
L71:
  addl   $0x1,-0x2c(%rbp)
L38:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L73
  mov    -0x24(%rbp),%eax
  mov    %eax,-0x2c(%rbp)
  jmp    L52
L30:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x10(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x2c(%rbp),%eax
  sub    -0x24(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x8(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x2c(%rbp)
L52:
  mov    -0x2c(%rbp),%eax
  cmp    -0x20(%rbp),%eax
  jle    L30
  mov    -0x20(%rbp),%eax
  sub    -0x24(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x8(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x8(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *longestPalindrome(char *s) {
  int len = strlen(s);
  char *x = (char *)(malloc(sizeof(char) * (len + 2)));
  char *ans = (char *)(malloc(sizeof(char) * (len + 2)));
  strcpy(x + 1, s);
  int i;
  x[0] = -1;
  int ma = 1;
  int l, r;
  l = r = 1;
  for (i = 1; x[i]; ++i) {
    int s = i, e = i;
    while (x[e + 1] == x[e])
      ++e;
    i = e;
    while (x[s - 1] == x[e + 1])
      --s, ++e;
    if (e - s + 1 > ma) {
      ma = e - s + 1;
      l = s;
      r = e;
    }
  }
  for (i = l; i <= r; i++) {
    ans[i - l] = x[i];
  }
  ans[r - l + 1] = '\0';
  free(x);
  return ans;
}
```