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Question

我正在使用概念编写一些通用软件，我想检查结构上是否存在带有（void）（int，int）签名的特定函数符号名称。为此，我正在考虑通过模板专门化来解决此问题，但我有点迷路。

如果您对概念不满意，我想做的就是工作，并且会遇到编译时错误：

struct TypeA {
  // Passes concept
  void process (int a ,int b) const {};
};

struct TypeB {
  // Does not pass concept
  void process (float a) const {};
};

struct TestConcepts {
  /* concept code here */
  TestConcepts(T t) {
    process_concept(t.process);
  };
};

int main(void) {
  // Should pass
  TestConcept(TypeA{});
  // Should throw error
  TestConcept(TypeB{});
  return 0;
}

我很难填补空白，但这是我到目前为止所要做的：

struct TestConcepts {
      /* concept code here */
      struct process_concept {
        process_concept((V*)(IA,IB)){
            if (is_integral<IA>::value && is_integral<IB>::value && is_same<V, void>) {
                return;
            }
            static_assert(false, "You must provide function called process of type (void)(int,int)");
        };
    };
      TestConcepts(T t) {
        process_concept(&t.process);
      };
    };

不幸的是，这不起作用。如何正确获得此功能签名？

Answer 1

如何使用返回已声明函数指针的函数？

struct TypeA {
    // Passes concept
    void process (int a ,int b) const {};
};

struct TypeB {
    // Does not pass concept
    void process (float a) const {};
};

template<typename T>
auto TestConcepts(T) -> void(T::*)(int, int) const
{
    return &T::process;
}

int main(void) {
    // Should pass
    TestConcepts(TypeA{});
    // Should throw error
    TestConcepts(TypeB{});
    return 0;
}

输出：

Error(s):

source_file.cpp: In instantiation of ‘void (T::* TestConcepts(T))(int, int) const [with T = TypeB]’:
source_file.cpp:26:23:   required from here
source_file.cpp:19:16: error: cannot convert ‘void (TypeB::*)(float) const’ to ‘void (TypeB::*)(int, int) const’ in return
     return &T::process;
                ^

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如果要像aschepler建议的那样包含void process(long int a, long int b) const;或void process(int a, int b, int c=0) const;，则可以使用类型特征。

struct TypeA {
    // Passes concept
    void process(int a, int b) const {};
};

struct TypeB {
    // Does not pass concept
    void process(float a) const {};
};

struct TypeC {
    // Passes concept
    void process(long int a, long int b) const {};
};

struct TypeD {
    // Passes concept
    void process(int a, int b, int c = 0) const {};
};

struct TypeE {
    // Does not pass concept
    void process(int a, int b, int c) const {};
};

#include <type_traits>
template<typename T, typename A1, typename A2, typename... An>
typename std::enable_if<
    std::is_integral<A1>::value &&
    std::is_integral<A2>::value
>::type
TestProcess(const T& t, void(T::*)(A1, A2, An...) const) {
    t.process(1, 2);
};

template<typename T>
void TestConcepts(const T& t)
{
    TestProcess(t, &T::process);
}

int main(void) {
    // Passes
    TestConcepts(TypeA{});
    // Throws compilation error
    TestConcepts(TypeB{});
    // Passes
    TestConcepts(TypeC{});
    // Passes
    TestConcepts(TypeD{});
    // Throws compilation error
    TestConcepts(TypeE{});

    return 0;
}

如何建立具有特定签名的功能概念？

1 个答案:

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