非类型参数模板函数，错误：必须调用对非静态成员函数的引用

编程问答 2022-05-02

问题描述

我正在尝试使用非类型参数模板函数作为类的成员，但遇到了错误。下面是一个最小的工作示例

#include <iostream>

enum Mode {ka,kb,kab};

class Foo {
  public:
    Foo(const double& x = 1.0,const Mode& y = ka) : x_(x),y_(y) {;}

    Mode get() const{
      return y_;
    }
    template<Mode mode> double FooFunc();

  private:
    double x_;
    Mode y_;
};

template<> double Foo::FooFunc<ka>() {
  return 1/x_;
}

template<> double Foo::FooFunc<kb>() {
  return 1/x_/x_;
}

template<> double Foo::FooFunc<kab>() {
  return 1/x_/x_/x_;
}

int main() {
  Foo foo1(2,ka),foo2(2,kb),foo3(2,kab);

  std::cout << "foo1 function result: " << foo1.FooFunc<foo1.get()> << "\n";
  std::cout << "foo2 function result: " << foo2.FooFunc<foo2.get()> << "\n";                                                                                                                                                                                                  
  std::cout << "foo3 function result: " << foo3.FooFunc<foo3.get()> << "\n";

  return 0;
}

我正在用 clang 编译它，如下c++ foo.cpp -o foo -O3 -std=c++11。这会导致以下错误，我不确定是什么问题。任何帮助将不胜感激。

foo.cpp:34:48: error: reference to non-static member function must be called
        std::cout << "foo1 function result: " << foo1.FooFunc<foo1.get()> << "\n";
                                                 ~~~~~^~~~~~~~~~~~~~~~~~~
foo.cpp:12:30: note: possible target for call
                template<Mode mode> double FooFunc();
                                           ^
foo.cpp:35:48: error: reference to non-static member function must be called
        std::cout << "foo2 function result: " << foo2.FooFunc<foo2.get()> << "\n";
                                                 ~~~~~^~~~~~~~~~~~~~~~~~~
foo.cpp:12:30: note: possible target for call
                template<Mode mode> double FooFunc();
                                           ^
foo.cpp:36:48: error: reference to non-static member function must be called
        std::cout << "foo3 function result: " << foo3.FooFunc<foo3.get()> << "\n";
                                                 ~~~~~^~~~~~~~~~~~~~~~~~~
foo.cpp:12:30: note: possible target for call
                template<Mode mode> double FooFunc();
                                           ^
3 errors generated.

解决方法

您的代码的问题在于模板参数必须是编译时常量。对于函数调用，这意味着它必须是 constexpr，而对于类方法，这意味着它必须是 static。

你只有两个选择：

  • 要么将 Mode 的配置留给运行时，并允许用户潜在地重新配置 Mode。为此，您必须在函数内引入 if/else 或 switch 语句。

    class Foo {
  public:
    constexpr Foo(double const x = 1.0,Mode const y = Mode::ka) noexcept
      : x_{x},y_{y} {
      return;
    }
    Mode get() const noexcept {
      return y_;
    }
    double FooFunc() const {
      switch (y_) {
        case Mode::ka: 
          return 1.0/x_;
        case Mode::kb:
          return 1.0/x_/x_;
        case Mode::kab:
          return 1.0/x_/x_/x_;
        default:
          // Throw error
      }
    }
  private:
    double x_;
    Mode y_;
};

  • 如果您不打算让用户随时更改 Mode，您可以将其设为类模板并专门化相应的功能。这样，虽然无法即时更改模式。

    template <Mode M = Mode::ka>
class Foo {
  public:
    constexpr Foo(double const x = 1.0) noexcept
      : x_(x) {
      return;
    }
    static constexpr Mode get() noexcept {
      return M;
    }
    double FooFunc() const noexcept;
  private:
    double x_;
    Mode y_;
};

template<>
inline double Foo<Mode::ka>::FooFunc() const noexcept {
  return 1.0/x_;
}
template<>
inline double Foo<Mode::kb>::FooFunc() const noexcept {
  return 1.0/x_/x_;
}
template<>
inline double Foo<Mode::kab>::FooFunc() const noexcept {
  return 1.0/x_/x_/x_;
}
,

使用 constexpr，您可以：

enum Mode {ka,kb,kab};

class Foo {
public:
    constexpr Foo(const double& x = 1.0,const Mode& y = ka) : x_(x),y_(y) {}

    constexpr Mode get() const{
      return y_;
    }
    template<Mode mode> constexpr double FooFunc() const;

private:
    double x_;
    Mode y_;
};

template<> constexpr double Foo::FooFunc<ka>() const {
  return 1/x_;
}

template<> constexpr double Foo::FooFunc<kb>() const {
  return 1/x_/x_;
}

template<> constexpr double Foo::FooFunc<kab>() const {
  return 1/x_/x_/x_;
}

int main() {
  constexpr Foo foo1(2,ka),foo2(2,kb),foo3(2,kab);

  std::cout << "foo1 function result: " << foo1.FooFunc<foo1.get()>() << "\n";
  std::cout << "foo2 function result: " << foo2.FooFunc<foo2.get()>() << "\n";
  std::cout << "foo3 function result: " << foo3.FooFunc<foo3.get()>() << "\n";
}

Demo

c++c++11templatestemplatestemplates