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

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  mov    $0x0,%eax
  call   <initMap@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <banner@plt>
  movl   $0x0,-0x1c(%rbp)
  movl   $0x5,-0x18(%rbp)
  jmp    L1
L90:
  lea    D78(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  lea    -0x1c(%rbp),%rax
  mov    %rax,%rsi
  lea    D42(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  test   %rax,%rax
  jg     L18
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  nop
L140:
  mov    $0x0,%eax
  call   <getchar@plt>
  cmp    $0xa,%al
  jne    L140
  addl   $0x1,-0x18(%rbp)
  jmp    L120
L18:
  mov    -0x1c(%rbp),%edx
  mov    -0x10(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <setMap@plt>
  test   %rax,%rax
  jne    L78
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <check@plt>
  test   %rax,%rax
  jne    L36
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <outMap@plt>
  lea    D17(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <setNPC@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <outMap@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <check@plt>
  test   %rax,%rax
  je     L120
  jmp    L28
L78:
  addl   $0x1,-0x18(%rbp)
L120:
  subl   $0x1,-0x18(%rbp)
L1:
  cmpl   $0x0,-0x18(%rbp)
  jne    L90
  jmp    L28
L36:
  nop
L28:
  lea    D62(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <check@plt>
  mov    %eax,-0x14(%rbp)
  cmpl   $0x0,-0x14(%rbp)
  jne    L146
  lea    D18(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L138
L146:
  cmpl   $0x1,-0x14(%rbp)
  jne    L155
  lea    D52(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
  jmp    L138
L155:
  cmpl   $0x2,-0x14(%rbp)
  jne    L138
  lea    D55(%rip),%rax
  mov    %rax,%rdi
  call   <printf@plt>
L138:
  mov    -0x10(%rbp),%rax
  mov    %rax,%rdi
  call   <free@plt>
  mov    -0x14(%rbp),%eax
  mov    %eax,%edi
  call   <writeRecord@plt>
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L31
  call   <__stack_chk_fail@plt>
L31:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: label: D78
data_type: string
name: disassemble_data
arguments: label: D42
data_type: string
name: disassemble_data
arguments: label: D26
data_type: string
name: disassemble_data
arguments: label: D17
data_type: string
name: disassemble_data
arguments: label: D62
data_type: string
name: disassemble_data
arguments: label: D18
data_type: string
name: disassemble_data
arguments: label: D52
data_type: string
name: disassemble_data
arguments: label: D55
data_type: string

tool: D78: Dein Zug: 

tool: D42: %d

tool: D26: 
Fehler bei der Eingabe!


tool: D17: Gegner ist am Zug:


tool: D62: Der Sieger ist...


tool: D18: Niemand


tool: D52: Der Spieler


tool: D55: Der Computer


assistant: ```c
int main(void) {
  char *map = initMap();
  banner(map);
  int input = 0;
  int turn;
  for (turn = 5; turn != 0; turn--) {
    printf("Dein Zug: ");
    if (scanf("%d", &input) <= 0) {
      printf("\nFehler bei der Eingabe!\n");
      while (getchar() != '\n')
        ;
      turn++;
      continue;
    }
    if (setMap(map, input) == 0) {
      if (check(map) != 0)
        break;
      outMap(map);
      printf("Gegner ist am Zug:\n");
      setNPC(map);
      outMap(map);
      if (check(map) != 0)
        break;
    } else {
      turn++;
    }
  }
  printf("Der Sieger ist...\n");
  int result = check(map);
  if (result == 0)
    printf("Niemand\n");
  else if (result == 1)
    printf("Der Spieler\n");
  else if (result == 2)
    printf("Der Computer\n");
  free(map);
  writeRecord(result);
  return 0;
}
```