模板函數作為模板函子的模板參數

Question

我想做這樣的事情，是否有人知道這是否可能？

template<typename T> using pLearn = T (*)(T, T, const HebbianConf<T> &);
template<typename T> using pNormal = T (*)(T, T);
template<typename T> using pDerivative = T (*)(T, T, T);

template <class Type, pLearn LearnCB, pNormal NormalCB, pDerivative DerivCB>
class TransfFunction {
public:
  static Type learn(Type a, Type b, const HebbianConf<Type> &setup) { return LearnCB<Type>(a, b, setup); };
  static Type normal(Type a, Type b) { return NormalCB<Type>(a, b); };
  static Type normal(Type a, Type b, Type c) { return DerivCB<Type>(a, b, c); };
};

錯誤：

In file included from /Functions.cpp:2:0:
/Functions.h:207:23: error: ‘pLearn’ is not a type
 template <class Type, pLearn LearnCB, pNormal NormalCB, pDerivative DerivCB>
                       ^
/Functions.h:207:39: error: ‘pNormal’ is not a type
 template <class Type, pLearn LearnCB, pNormal NormalCB, pDerivative DerivCB>
                                       ^
/Functions.h:207:57: error: ‘pDerivative’ is not a type
 template <class Type, pLearn LearnCB, pNormal NormalCB, pDerivative DerivCB>

Answer 1

錯誤說明了一切：

In file included from /Functions.cpp:2:0:
/Functions.h:207:23: error: ‘pLearn’ is not a type
 template <class Type, pLearn LearnCB, pNormal NormalCB, pDerivative DerivCB>
                       ^

pLearn不是類型pLearn是別名模板。 模板非類型參數需要type 。 您需要為其提供類型參數。 其他兩個相同：

template <class Type,
          pLearn<Type> LearnCB,
          pNormal<Type> NormalCB,
          pDerivative<Type> DerivCB>
class TransfFunction { ... };