user: What would the C source look like for the assembly code provided here?

```asm
<trans>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L7
L43:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  mov    $0x21,%eax
  sub    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    $0x21,%eax
  sub    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
L7:
  cmpl   $0x10,-0xc(%rbp)
  jle    L43
  movl   $0x0,-0xc(%rbp)
  jmp    L67
L31:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  movl   $0xd,-0x4(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L22
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jg     L22
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x41,(%rax)
  movsbl -0xd(%rbp),%eax
  sub    $0x41,%eax
  mov    %eax,-0x8(%rbp)
  jmp    L8
L22:
  movsbl -0xd(%rbp),%eax
  sub    $0x61,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x61,(%rax)
L8:
  mov    -0x8(%rbp),%edx
  mov    -0x4(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  imul   $0x4ec4ec4f,%rdx,%rdx
  shr    $0x20,%rdx
  sar    $0x3,%edx
  mov    %eax,%ecx
  sar    $0x1f,%ecx
  sub    %ecx,%edx
  mov    %edx,-0x8(%rbp)
  mov    -0x8(%rbp),%edx
  imul   $0x1a,%edx,%edx
  sub    %edx,%eax
  mov    %eax,-0x8(%rbp)
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %eax,%edx
  mov    -0x8(%rbp),%eax
  lea    (%rdx,%rax,1),%ecx
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  addl   $0x1,-0xc(%rbp)
L67:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L31
  mov    -0x18(%rbp),%rax
  mov    %rax,%rsi
  lea    D7(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: label: D7
type_infered: string

tool: D7: %s


assistant: ```c
void trans(char arr[34]) {
  int i;
  char ch;
  for (i = 0; i < 34 / 2; i++) {
    ch = arr[i];
    arr[i] = arr[33 - i];
    arr[33 - i] = ch;
  }
  for (i = 0; arr[i] != '\0'; i++) {
    ch = arr[i];
    int number, n = 13;
    if ((arr[i] >= 'A') && (arr[i] <= 'Z')) {
      arr[i] = 'A';
      number = ch - 'A';
    } else {
      number = ch - 'a';
      arr[i] = 'a';
    }
    number = (number + n) % 26;
    arr[i] = arr[i] + number;
  }
  printf("%s\n", arr);
}
```