20. STL deque2

# assign & swap

- assign또한 vector에서의 assign과 동일하다.

- 값을 지우고 다시 넣는다고 생각하면 되겠다.

- deque1.assign(5,6) // 6을 5번 채운다.

- deque2.assign(deque1.begin(), deque1.end()) // deque1의 내용을 복사해서 채운다.

- swap은 값을 바꾼다.(사이즈 달라도 가능)

#include <iostream>
#include <deque>
using namespace std;
 
void print(deque<int> deque1) {
  for(deque<int>::iterator iterPos = deque1.begin(); iterPos != deque1.end(); ++iterPos) {
    cout << "deque : " << *iterPos << endl;
  }
}
 
int main() {
  deque<int> deque1;
  cout << "assign 1" << endl;
  deque1.assign(5, 6);
  print(deque1);
  
  cout << endl << "assign 2" << endl;
  deque<int> deque2;
  deque2.assign(deque1.begin(), deque1.end());
  print(deque2);
  
  // swap
  deque<int> deque3;
  deque3.push_back(10);
  deque3.push_back(20);
  deque3.push_back(30);
  deque3.swap(deque1);
  cout << endl << "swap" << endl;
  print(deque1);
  cout << endl;
  print(deque3);
  
  return 0;
}

* 결과

assign 1
deque : 6
deque : 6
deque : 6
deque : 6
deque : 6

assign 2
deque : 6
deque : 6
deque : 6
deque : 6
deque : 6

swap
deque : 10
deque : 20
deque : 30
deque : 6
deque : 6
deque : 6
deque : 6
deque : 6

* 마지막으로 책에선 또 과제를 내어준다.

1. deque을 이용해 FIFO, LIFO 스택을 구현해라.

#include <iostream>
#include <deque>
#include <string.h>
using namespace std;
 
template <typename T>
class Stack{
  public :
    Stack() { Clear(); }
    void Clear() { Elements.clear(); }
    bool IsEmpty() const { return true == Elements.empty() ? true : false;  }
    void SetLIFO(bool LIFO) { m_bLIFO = LIFO; }
    bool push(T data) { Elements.push_back(data); }
    bool pop(T* Value) {
      if(IsEmpty()) return false;
      if(m_bLIFO) {
        memcpy(Value, &Elements[Elements.size()-1], sizeof(T));
        Elements.pop_back();
      }
      else {
        memcpy(Value, &Elements[0], sizeof(T));
        Elements.pop_front();
      }
      return true;
    }
    void print() const { for(int i = 0; i < Elements.size(); ++i) cout << "Stack : " << Elements[i] << endl; }
  private :
    deque<T> Elements;
    bool m_bLIFO;
};
 
int main() {
  Stack<int> intStack;
  intStack.push(100);
  intStack.push(200);
  intStack.push(300);
  intStack.SetLIFO(true);
  int Value;
  intStack.pop(&Value);
  cout << "Value : " << Value << endl;
  intStack.SetLIFO(false);
  intStack.pop(&Value);
  cout << "Value : " << Value << endl;
  
  return 0;
}

# 결과

Value : 300
Value : 100

2. deque을 사용해서 Undo, Redo를 구현하라.

Undo - deque에 숫자를 저장하고 Undo를 하면 마지막에 넣은 데이터를 뺀다.

Redo - 뺀 데이터를 다시 넣는다.

#include <iostream>
#include <deque>
using namespace std;
 
deque<int> back_deque;
 
void Undo(deque<int>* Deque) {
  back_deque.push_back(*(Deque->end()-1));
  Deque->pop_back();
}
 
void Redo(deque<int>* Deque) {
  Deque->push_back(*(back_deque.end()-1));
  back_deque.pop_back();
}
 
void print(deque<int> Deque) {
  for(deque<int>::iterator iterPos = Deque.begin(); iterPos!=Deque.end(); ++iterPos) {
    cout << "Deque : " << *iterPos << endl;
  }
}
 
int main() {
  deque<int> deque1;
  deque1.push_back(100);
  deque1.push_back(200);
  deque1.push_back(300);
  deque1.push_back(400);
  cout << "Undo Test" << endl;
  Undo(&deque1);
  cout << endl << "deque1" << endl;
  print(deque1);
  cout << endl << "back_deque" << endl;
  print(back_deque);
  
  cout << endl << "Redo Test "<< endl;
  Redo(&deque1);
  cout << endl << "deque1" << endl;
  print(deque1);
  cout << endl;
  cout << "back_deque" << endl;
  print(back_deque);
  return 0;
}

# 결과

Undo Test

deque1
Deque : 100
Deque : 200
Deque : 300

back_deque
Deque : 400

Redo Test 

deque1
Deque : 100
Deque : 200
Deque : 300
Deque : 400

back_deque

* 참고 : Thinking About C++ STL Programming.