Question

我想在用户端实现什么语法：

double a(1.), b(2.), deps(.1);
bool res1 = compare<double>()(a, b);        // works with default eps
bool res2 = compare<double, &deps>()(a, b);  // works with explicitly provided eps
float c(1.), d(1.). feps(.1);
bool res3 = compare<float>()(c, d);  // don't have default eps - must not compile
bool res4 = compare<float, &feps>()(c, d);   // works only with provided eps

我现在有什么实现（不工作，因为不允许使用默认参数进行部分特化）：

extern double eps_double; // somewhere defined and initialized

template<typename T, const T* eps>
struct compare { // actually inherits std::binary_function
  bool operator()(const T& t1, const T& t2) {
    return t1 < t2 - *eps;
  }
};
template<const double* eps = &eps_double>
struct compare<double, eps> { // the same as in default implementation
};

我尝试过使用静态成员的enable_if和包装类，但不能将静态成员分配给外部变量;

更新实际问题是一般结构和专用结构的名称相等。我不知道如何在没有重命名的情况下使其工作：

// treats all explicitly passed eps and don't need default parameter
template<typename T, const T* eps>
struct compare_eps { // need another name! 
  bool operator()(const T& t1, const T& t2) {
    return t1 < t2 - *eps;
  }
};
// don't need default parameter either
// because we always know what eps to use for a concrete type
template<typename T>
struct compare { 
  // define nothing -> will not compile on types we don't have specialization for
}; 
template<>
struct compare<double> { 
  // use here eps_double hardcoded
};

Answer 1

我不知道为什么你认为这是有道理的

compare<double, deps>

你无法做到这一点：模板参数不能是double类型的值（它们可以是double类型的左值，但你的模板需要double的地址，所以那是关闭的）。

您可以使用函数模板使语法正常工作

extern double eps_double;

template<typename T>
types::compare<T, &eps_double> compare(
  typename enable_if<is_same<T, double>>::type * = 0
) {
  return types::compare<T, &eps_double>(); 
}

template<typename T, const T *eps>
types::compare<T, eps> compare() {
  return types::compare<T, eps>(); 
}

或者，如果你想要一些丑陋的黑客，你可以使用类模板

template<typename T, const T* eps = &eps_double>
struct compare { 
  bool operator()(const T& t1, const T& t2) {
    return t1 < t2 - *eps;
  }
};

如果同时提供两个参数，则不会使用默认参数。如果仅提供<double>，则将使用默认参数并将起作用。如果您只提供<float>，则也会使用默认参数，但不起作用。

Answer 2

您需要更改拥有比较运算符的结构，以便您可以专门化外部模板，请参阅：http://ideone.com/xqtjz

代码是：

extern double eps_double; // somewhere defined and initialized
extern double deps;  // NOTE: you have to make these extern a well, else cannot work
extern float feps;

template<typename T>
struct compare {
  // this internal structure now has the operator()
  template <const T* eps>
  struct it
  {
    bool operator()(const T& t1, const T& t2) const {
      return t1 < t2 - *eps;
    }
  };
};

// specialize compare for double
template<>
struct compare<double>
{
  // here you can provide the default
  template<const double* eps=&eps_double>
  struct it
  {
    bool operator()(const double& t1, const double& t2)
    {
      return t1 < t2 - *eps;
    }
  };
};

int main(void)
{
  double a(1.), b(2.);
  bool res1 = compare<double>::it<>()(a, b);        // works with default eps
  bool res2 = compare<double>::it<&deps>()(a, b);   // works with explicitly provided eps
  float c(1.), d(1.);
  bool res3 = compare<float>::it<>()(c, d);  // don't have default eps - will not compile
  bool res4 = compare<float>::it<&feps>()(c, d);   // works only with provided eps
}

Answer 3

我会选择类似特质的方法：

template<class T>
struct DefaultEps;

template<>
struct DefaultEps<double>
{
  static const double eps = 4.0;
};

// may or may not be needed in .cpp
// const double DefaultEps<double>::eps;

template<>
struct DefaultEps<float>
{
  static const float eps = 4.0;
};

// may or may not be needed in .cpp
// const float DefaultEps<float>::eps;

template<class T, class E = DefaultEps<T> >
struct Compare
{
  bool operator()(T const &t1, T const &t2)
  {
    return(t1 < t2 - E::eps);
  }
};

然后当需要特定的epsilon时：

struct SpecificEps
{
  static const float eps = 4.0;
};

// may or may not be needed in .cpp
// const float SpecificEps::eps;

并使用它：

Compare<float, SpecificEps>()(...);

部分特化的默认参数

3 个答案: