我正在實現A *算法來解決一些問題--8拼圖問題和另一個問題。對於一個明星,我實現了在A_star.hpp常見三種:對繼承函數的未定義引用
template <class U>
class Heuristic{
public:
virtual int getHeuristic(Node<U> currNode, Node<U> target);
};
template <class V>
class NextNodeGenerator{
public:
virtual vector<pair<V, int> > generate(Node<V> curr);
};
template <class W>
class CompareVal{
public:
virtual bool compare(W val1, W val2);
};
爲解決8數碼問題,我實現了三個子類在prob2.cpp上述每個泛型類:
template <class hT>
class PuzzleHeuristic: public Heuristic<hT>{
public:
virtual int getHeuristic(Node<hT> currNode, Node<hT> target){
//Code for getHeuristic
}
};
template <class cT>
class PuzzleCompareVal: public CompareVal<cT>{
public:
virtual bool compare(cT val1, cT val2){
//Code for compare
}
};
template <class nT>
class PuzzleNNG: public NextNodeGenerator<nT>{
public:
virtual vector<pair<nT, int> > generate(Node<nT> curr){
//Code for generate
}
在A_star.hpp,我也有一個愛仕達類:
template <class Y>
class AStar{
Heuristic<Y> *h;
NextNodeGenerator<Y> *nng;
CompareVal<Y> *comp;
public:
void setHeuristic(Heuristic<Y> *hParam){
h = hParam;
}
void setNNG(NextNodeGenerator<Y> *nngParam){
nng = nngParam;
}
void setCompareVal(CompareVal<Y> *compParam){
comp = compParam;
}
vector<Node<Y> > solve(Y start, Y target){
//Code for solve
}
在prob2.cpp我的main()函數,我創造了一個愛仕達對象(數組是我已經定義的模板類分開):
int main()
{
PuzzleHeuristic<Array<int> > pH;
PuzzleCompareVal<Array<int> > pCV;
PuzzleNNG<Array<int> > pNNG;
AStar<Array<int> > aStar;
aStar.setHeuristic(&pH);
aStar.setNNG(&pNNG);
aStar.setCompareVal(&pCV);
vector<Node<Array<int> > > answer = aStar.solve(start, target);
}
在編譯時,我得到了以下錯誤:
/tmp/ccCLm8Gn.o:(.rodata._ZTV17NextNodeGeneratorI5ArrayIiEE[_ZTV17NextNodeGeneratorI5ArrayIiEE]+0x10): undefined reference to
NextNodeGenerator<Array<int> >::generate(Node<Array<int> >)' /tmp/ccCLm8Gn.o:(.rodata._ZTV10CompareValI5ArrayIiEE[_ZTV10CompareValI5ArrayIiEE]+0x10): undefined reference to
CompareVal >::compare(Array, Array)' /tmp/ccCLm8Gn.o:(.rodata._ZTV9HeuristicI5ArrayIiEE[_ZTV9HeuristicI5ArrayIiEE]+0x10): undefined reference to `Heuristic >::getHeuristic(Node >, Node >)' collect2: error: ld returned 1 exit status Blockquote
我懷疑問題是由於模板函數的繼承。什麼可能導致錯誤?