user: 请从给定的汇编指令推导出C源代码：

```asm
<submenu>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  mov    %rsi,%r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %edx,0x8(%rsp)
  mov    %ecx,0xc(%rsp)
  mov    %rdi,(%rsp)
  lea    D54(%rip),%rdi
  mov    %fs:0x28,%rax
  mov    %rax,0x18(%rsp)
  xor    %eax,%eax
  call   <puts@plt>
  lea    D23(%rip),%rdi
  call   <puts@plt>
  lea    D36(%rip),%rdi
  call   <puts@plt>
  lea    D12(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  lea    0x14(%rsp),%rsi
  lea    D33(%rip),%rdi
  call   <scanf@plt>
  mov    0x14(%rsp),%eax
  cmp    $0x1,%eax
  je     L65
  cmp    $0x2,%eax
  je     L19
L26:
  test   %eax,%eax
  je     L22
L0:
  mov    0x18(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L24
  add    $0x28,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax)
L65:
  mov    0x8(%rsp),%edx
  test   %edx,%edx
  jle    L0
  mov    (%rsp),%rbx
  xor    %ebp,%ebp
  xor    %r14d,%r14d
  lea    D46(%rip),%r13
L47:
  mov    (%rbx),%edx
  add    $0x1,%r14d
  lea    D58(%rip),%rdi
  xor    %eax,%eax
  mov    %r14d,%esi
  call   <printf@plt>
  mov    (%rbx),%eax
  test   %eax,%eax
  jle    L45
  lea    0x1(%rbp),%edx
  movslq %edx,%r15
  nopl   0x0(%rax,%rax,1)
L82:
  mov    -0x4(%r12,%r15,4),%esi
  mov    %r13,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    %r15d,%ecx
  add    $0x1,%r15
  mov    %r15d,%eax
  sub    %ebp,%eax
  sub    $0x1,%eax
  cmp    %eax,(%rbx)
  jg     L82
  mov    %ecx,%ebp
L45:
  lea    D64(%rip),%rdi
  add    $0x4,%rbx
  call   <puts@plt>
  cmp    0x8(%rsp),%r14d
  jne    L47
  mov    0x14(%rsp),%eax
  cmp    $0x2,%eax
  jne    L26
L19:
  mov    0xc(%rsp),%ecx
  mov    0x8(%rsp),%edx
  mov    %r12,%rsi
  mov    (%rsp),%rdi
  call   <euleriano@plt>
  mov    0x14(%rsp),%eax
  test   %eax,%eax
  jne    L0
L22:
  call   <limpa@plt>
  jmp    L0
L24:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: unpack_data
arguments: data_label: D54
type_infered: string
name: unpack_data
arguments: data_label: D23
type_infered: string
name: unpack_data
arguments: data_label: D36
type_infered: string
name: unpack_data
arguments: data_label: D12
type_infered: string
name: unpack_data
arguments: data_label: D33
type_infered: string
name: unpack_data
arguments: data_label: D46
type_infered: string
name: unpack_data
arguments: data_label: D58
type_infered: string
name: unpack_data
arguments: data_label: D64
type_infered: string

tool: D54: Digite 1 para mostrar o grafo


tool: D23: Digite 2 para encontrar um caminho euleriano


tool: D36: Digite 0 para retornar ao menu anterior


tool: D12: ::>

tool: D33: %i

tool: D46: %i 

tool: D58: Indice:%i Grau: %i Vizinhos: 

tool: D64: 


assistant: ```c
void submenu(int *vert, int *adj, int cont, int cont2) {
  int opcao, i, j, k = 0;
  puts("Digite 1 para mostrar o grafo\n");
  puts("Digite 2 para encontrar um caminho euleriano\n");
  puts("Digite 0 para retornar ao menu anterior\n");
  printf("::>");
  scanf("%i", &opcao);
  if (opcao == 1) {
    for (i = 0; i < cont; i++) {
      printf("Indice:%i Grau: %i Vizinhos: ", i + 1, vert[i]);
      for (j = 0; j < vert[i]; j++) {
        printf("%i ", adj[k]);
        k++;
      }
      puts("\n");
    }
  }
  if (opcao == 2) {
    euleriano(vert, adj, cont, cont2);
  }
  if (opcao == 0) {
    limpa();
    return;
  }
}
```