C++模板函数从可调用类实例到std::function

huangapple
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英文:

C++ template function from callable class instance to std::function

问题

I do need std::function for stateful operation.

So I would like to make an instance of a class implementing operator() into std::function. I did test std::bind works, but can't wrap it up as a function.

  1. struct StatefulOperation
  2. {
  3.     StatefulOperation(int num) : num_(num) {}
  4.     int operator()(int i)
  5.     {
  6.         num_ += i;
  7.         return num_;
  8.     }
  9.     
  10.     int num_;
  11. };
  14. // This works
  16. const StatefulOperation obj(40);
  17.     
  18. using std::placeholders::_1;
  19. std::function<int(int)> op = std::bind(&StatefulOperation::operator(), obj, _1);
  22. // Want something like this
  24. template<typename A, typename B, typename C>
  25. std::function<B(A)> op(C& obj);
  28. // This doesn't work
  30. template<typename A, typename B, typename C>
  31. std::function<B(A)> op(C& obj)
  32. {
  33.     using std::placeholders::_1;
  34.     return std::bind(&C::operator(), obj, _1);
  35. }

The last implementation attempt fails with an error like below:

  1. main.cpp:52:21: note:   template argument deduction/substitution failed:
  2. main.cpp:75:35: note:   couldn't deduce template parameter 'A'
  3.    75 |     std::function<int(int)> f = op(a);
  4.       |                                 ~~^~~

How can I get this done?

英文:

I do need std::function for stateful operation.

So I would like to make an instance of a class implementing operator() into std::function. I did test std::bind works, but can't wrap it up as a function.

  1. struct StatefulOperation
  2. {
  3.     StatefulOperation(int num) : num_(num) {}
  4.     int operator()(int i)
  5.     {
  6.         num_ += i;
  7.         return num_;
  8.     }
  9.     
  10.     int num_;
  11. };
  14. // This works
  16. const StatefulOperation obj(40);
  17.     
  18. using std::placeholders::_1;
  19. std::function&lt;int(int)&gt; op = std::bind(&amp;StatefulOperation::operator(), obj, _1);
  22. // Want something like this
  24. template&lt;typename A, typename B, typename C&gt;
  25. std::function&lt;B(A)&gt; op(C&amp; obj);
  28. // This doesn&#39;t work
  30. template&lt;typename A, typename B, typename C&gt;
  31. std::function&lt;B(A)&gt; op(C&amp; obj)
  32. {
  33.     using std::placeholders::_1;
  34.     return std::bind(&amp;C::operator(), obj, _1);
  35. }

The last implementation attempt fails with error like below;

  1. main.cpp:52:21: note:   template argument deduction/substitution failed:
  2. main.cpp:75:35: note:   couldn’t deduce template parameter ‘A’
  3.    75 |     std::function&lt;int(int)&gt; f = op(a);
  4.       |                                 ~~^~~

How can I get this done?

答案1

得分: 2

编译器无法仅从返回类型中推断出A和B，您需要指定它们：

  1. std::function<int(int)> f = op<int, int>(a);

然而，您实际上不需要这些额外的模板参数。只需使用auto来推断返回类型：

  1. template<typename C>
  2. auto op(C& obj)
  3. {
  4.     using std::placeholders::_1;
  5.     return std::bind(&C::operator(), std::ref(obj), _1);
  6. }

返回类型不会完全相同，但可以转换为您想要的std::function<B(A)>。如果您想指定确切的返回类型，可以结合使用this和std::invoke_result_t来实现。

英文:

The compiler cannot deduce A and B just from the return type. You need to specify them:

  1. std::function&lt;int(int)&gt; f = op&lt;int, int&gt;(a);

However, you don't really need these additional template parameters. Just use auto to deduce the return type:

  1. template&lt;typename C&gt;
  2. auto op(C&amp; obj)
  3. {
  4.     using std::placeholders::_1;
  5.     return std::bind(&amp;C::operator(), std::ref(obj), _1);
  6. }

The return type will not be exactly the same, but convertible to what you want (std::function&lt;B(A)&gt;). If you want to specify the exact return type, you can do so with a combination of this and std::invoke_result_t.

答案2

得分: 1

Instead of using bind you should use lambda expressions. Also you StatefulOperation should NOT be const (otherwise it can't update its internal state).

  1. #include <functional>
  2. #include <iostream>
  4. struct StatefulOperation
  5. {
  6.     StatefulOperation(int num) : num_(num) {}
  7.     int operator()(int i)
  8.     {
  9.         num_ += i;
  10.         return num_;
  11.     }
  13.     int num_;
  14. };
  16. int main()
  17. {
  18.     /*const*/ StatefulOperation obj(40); // your state cannot be const since you are changing its internal state
  20.     // instead of bind rely on lambda expressions (bind is still there because of well... history)
  21.     std::function<int(int)> op{ [&](int num) { return obj(num); } };
  23.     for (int i = 0; i < 10; ++i)
  24.     {
  25.         std::cout << op(i) << " ";
  26.     }
  28.     return 0;
  29. }
英文:

Instead of using bind you should use lambda expressions. Also you StatefulOperation should NOT be const (otherwise it can't update its internal state).

  1. #include &lt;functional&gt;
  2. #include &lt;iostream&gt;
  4. struct StatefulOperation
  5. {
  6.     StatefulOperation(int num) : num_(num) {}
  7.     int operator()(int i)
  8.     {
  9.         num_ += i;
  10.         return num_;
  11.     }
  13.     int num_;
  14. };
  16. int main()
  17. {
  18.     /*const*/ StatefulOperation obj(40); // your state cannot be const since you are changing its internal state
  19.     
  20.     // instead of bind rely on lambda expressions (bind is still there because of well... history)
  21.     std::function&lt;int(int)&gt; op{ [&amp;](int num) { return obj(num); } };
  23.     for (int i = 0; i &lt; 10; ++i)
  24.     {
  25.         std::cout &lt;&lt; op(i) &lt;&lt; &quot; &quot;;
  26.     }
  28.     return 0;
  29. }

答案3

得分: 0

你可以用最简单的方式初始化一个std::function，因为你的类是可调用的，实现了一个具有期望类型的函数调用运算符：

  1. std::function<int(int)> op{StatefulOperation(40)};

不要忘记，当复制op时，状态也会被复制。

英文:

Since your class is a callable that implements a function call operator with the expected types, you can initialize a std::function in the most simple way:

  1. std::function&lt;int(int)&gt; op{StatefulOperation(40)};

Do not forget that the state is copied when op is copied.

huangapple
  • 本文由 发表于 2023年7月4日 20:43:55
  • 转载请务必保留本文链接：https://go.coder-hub.com/76612778.html
  • c++
  • c++11
  • std-function
  • stdbind
  • templates
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