Question

int listRecSearch(list<int>list, const int data)
{
  if (list.empty())
  {
    cout << "The number is not in the list, try again..." << endl;
    return -1;
  }
  else if (list.back() == data)
  {
//    cout << "list.size(): " << list.size() << endl;
    list.pop_back();//I needed the index begins from 0 instead of 1
    return list.size();
  }
  else
  {
//    cout << "list.back(): " << list.back() << endl;
    list.pop_back();
    listRecSearch(list, data);
  }
}

//funtion used
int main()
{
 list<int>list = listGenerator(size);//generate a list with 20 random numbers.
 cout << "Specify the element to be searched for: ";
 cin >> data;
 int position = listRecSearch(list, data);
 if (position > -1)
    cout << "\nFind the element at position: " << position << endl;
}

函数listRecSearch能够显示正确的list.size()值并更正pop_back值。但是一旦它返回，它总是返回一个垃圾值。我想在返回后仍然有一些步骤，但我不知道在哪里以及如何。

Answer 1

存在不返回值的代码路径。 listRecSearch(list, data);应该成为return listRecSearch(list, data);。

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