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我已經構建了一個旨在檢查二叉樹的迭代器類。這個迭代器波紋管所：C++在迭代器類中定義析構函數？
template <class Key> class intervalST_const_iterator
{
//friend class IntervalST<Key>;
private:
const Interval<Key> *interval;// equivalent to Interval<Key> const *interval; meaning: interval is apointer to the const object: Interval<Key>
//meaning that the object cannot be changed via interval
//interval can change here
public:
intervalST_const_iterator(const Interval<Key> *p): interval(p){}
//~intervalST_const_iterator() {delete interval;}
//the const at the end of this operator means that the operator will not change *this
bool operator != (const intervalST_const_iterator & other) const
{
return this->interval != other.interval;
}
bool operator == (const intervalST_const_iterator & other) const
{
return this->interval == other.interval;
}
//the const at the beginning means that the it's not allowed to change the object
const Interval<Key> operator *() const
{
return *(this->interval);
}
intervalST_const_iterator<Key> & left()
{
interval = interval->left;
return *this;
}
intervalST_const_iterator<Key> & right()
{
interval = interval->right;
return *this;
}
};
在這個類中，我還定義了一個析構函數~intervalST_const_iterator() {delete interval;}我已經把它放在評論。
爲了一棵樹類，看起來像重複我使用這個迭代器：
template <class Key> class intervalST_const_iterator;
template <class Key> class IntervalST
{
private:
Interval<Key> *root;
//friend class intervalST_const_iterator<Key>;//allow the iterator class to access the private section of intervalST
bool isRed(Interval<Key> *interval);
Interval<Key> *rotateLeft(Interval<Key> *h);
Interval<Key> *rotateRight(Interval<Key> *h);
Interval<Key> *put(Interval<Key> *h,Key lo, Key hi, Key val);
Interval<Key> *moveRedLeft(Interval<Key> *h);
Interval<Key> *moveRedRight(Interval<Key> *h);
Interval<Key> *deleteMin(Interval<Key> *h, Key hi);
Interval<Key> *balance(Interval<Key> *h);
Interval<Key> *remove(Interval<Key> *h, Key lo, Key hi);
Interval<Key> *min(Interval<Key> *h);
Interval<Key> *addDuplicate(Interval<Key> *h, Key hi);
Interval<Key> *removeDuplicate(Interval<Key> *h, Key low, Key hi);
Interval<Key> *getPointerToKey(Key low);
void flipColors(Interval<Key> *h);
void destroy(Interval<Key> *h);
void printTree(Interval<Key> *h, int indent);
Key maxVal(Interval<Key> *h);
int size(Interval<Key> *h);
bool isBST(Interval<Key> *x, Key min, Key max);
inline bool isBST(){return isBST(root,0,0);}
bool isSizeConsistent(Interval<Key> *x);
inline bool isSizeConsistent(){return isSizeConsistent(root);}
bool is23(Interval<Key> *x);
inline bool is23(){return is23(root);}
bool isBalanced();
bool isBalanced(Interval<Key> *x,int black);
int getKeySize(Key low);
int compare(Key a, Key b);
public:
//don't forget to build the constructor
//and overload the =equal operator
typedef intervalST_const_iterator<Key> const_iterator;//const iterator on a tree
const_iterator begin() const;
const_iterator end() const;
IntervalST():root(NULL){};
~IntervalST();
void remove(Key lo, Key hi);
void put(Key lo, Key hi);
inline int size(){return size(root);}
inline bool isEmpty(){return root == NULL;}
void print(int indent = 0);
void check();
};
相應begin()和end()方法迭代
template <typename Key> typename IntervalST<Key>::const_iterator IntervalST<Key>::begin() const
{
return const_iterator(root);
}
template <typename Key> typename IntervalST<Key>::const_iterator IntervalST<Key>::end() const
{
//return const_iterator(NULL,this);
return const_iterator(NULL);
}
然後我使用函數 - 請參閱波紋管 - 當迭代器類的析構函數未定義時運行良好，當析構函數定義爲時，會產生奇怪的結果。
爲什麼發生這種情況？ 我是否需要爲迭代器類定義析構函數以防止泄漏？ 如何來處理這個問題？ 感謝您的幫助。
template <typename Key> std::vector<Segment<Key> > findIntersections(const IntervalST<Key> &tree ,Segment<Key> segment)
{
//initialize a const iterator on the tree
intervalST_const_iterator<Key> iterator = tree.begin();
vector<Segment<Key> > intersections;
Key k = (*iterator).max;
//walk the tree
while(iterator != tree.end())
{
//if(*(iterator).intersects(segment.gety1(),segment.gety2())) intersections.push_back(segment);
if(intersects((*iterator).low,(*iterator).back,segment.gety1(),segment.gety2()))
{
intersections.push_back(segment);
return intersections;
}
else if (iterator.left() == tree.end()) iterator = iterator.right();
else if (segment.gety1() > (*iterator.left()).max) iterator = iterator.right();
else iterator = iterator.left();
}
return intersections;
}
您可能想了解[三項規則]（https://en.wikipedia.org/wiki/Rule_of_three_%28C%2B%2B_programming%29）。 –
@Joachim Pileborg我認爲問題**沒有鏈接到**我沒有定義一個副本ctor，並且我沒有重載equal（=）運算符。 – PhonoDots
@PhonoDots：確實;這是因爲你刪除了一些你沒有「新」的東西。不要這樣做。 （但是，如果它確實需要析構函數，那麼「三定律」將非常重要）。 –