c – 绑定移动结构功能

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我需要将结构与已删除的副本构造函数绑定到一个函数.我把我想要实现的目标减少到如下几个最小的例子中:
struct Bar {
    int i;
    Bar() = default;
    Bar(Bar&&) = default;
    Bar(const Bar&) = delete;
    Bar& operator=(const Bar&) = delete;
};

void foo(Bar b) {
    std::cout << b.i << std::endl;
}

int main()
{
    Bar b;
    b.i = 10;

    std::function<void()> a = std::bind(foo,std::move(b)); // ERROR
    a();

    return 0;
}

从编译器我只会哭泣和咬牙:

test.cpp:22:27: error: no viable conversion from 'typename _Bind_helper<__is_socketlike<void (&)(Bar)>::value,void (&)(Bar),Bar>::type' (aka '_Bind<__func_type (typename decay<Bar>::type)>') to 'std::function<void ()>'
    std::function<void()> a = std::bind(foo,std::move(b));
                          ^   ~~~~~~~~~~~~~~~~~~~~~~~~~~~~
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.1.0/../../../../include/c++/5.1.0/functional:2013:7: note: candidate constructor not viable: no known conversion from 'typename _Bind_helper<__is_socketlike<void (&)(Bar)>::value,Bar>::type' (aka '_Bind<__func_type (typename decay<Bar>::type)>') to 'nullptr_t' for 1st argument
      function(nullptr_t) noexcept
      ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.1.0/../../../../include/c++/5.1.0/functional:2024:7: note: candidate constructor not viable: no known conversion from 'typename _Bind_helper<__is_socketlike<void (&)(Bar)>::value,Bar>::type' (aka '_Bind<__func_type (typename decay<Bar>::type)>') to 'const std::function<void ()> &' for 1st argument
      function(const function& __x);
      ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.1.0/../../../../include/c++/5.1.0/functional:2033:7: note: candidate constructor not viable: no known conversion from 'typename _Bind_helper<__is_socketlike<void (&)(Bar)>::value,Bar>::type' (aka '_Bind<__func_type (typename decay<Bar>::type)>') to 'std::function<void ()> &&' for 1st argument
      function(function&& __x) : _Function_base()
      ^
/usr/bin/../lib/gcc/x86_64-linux-gnu/5.1.0/../../../../include/c++/5.1.0/functional:2058:2: note: candidate template ignored: substitution failure [with _Functor = std::_Bind<void (*(Bar))(Bar)>]: no matching function for call to object of
      type 'std::_Bind<void (*(Bar))(Bar)>'
        function(_Functor);
        ^
1 error generated.

所以我想询问是否有任何解决方法,允许我绑定到foo,同时保持Bar移动.

编辑:
还要考虑以下代码,其中变量b的生命在a被调用之前终止:

int main()
{
    std::function<void()> a;
    {
        Bar b;
        b.i = 10;
        a = std::bind(foo,std::move(b)); // ERROR
    }
    a();

    return 0;
}

解决方法

std :: function不能采取move-only调用.它将传入的类型删除,以调用(带签名),destroy和copy.1

只写一个move-only std ::函数只是一点点工作. Here is a stab at it在不同的上下文中. live example.

std :: packed_task有趣的是一个移动型的橡皮擦调用者,但它比你想要的更重,并且获得的价值是一种痛苦.

一个更简单的解决方案是滥用一个共享指针:

template<class F>
auto shared_function( F&& f ) {
  auto pf = std::make_shared<std::decay_t<F>>(std::forward<F>(f));
  return [pf](auto&&... args){
    return (*pf)(decltype(args)(args)...);
  };
}

它将一些可调用对象包装到共享指针中,将其置于lambda完美转发lambda中.

这说明了一个问题 – 呼叫不起作用!所有上述都有一个const调用.

你想要的是一个只能打一次的任务.

template<class Sig>
struct task_once;

namespace details_task_once {
  template<class Sig>
  struct ipimpl;
  template<class R,class...Args>
  struct ipimpl<R(Args...)> {
    virtual ~ipimpl() {}
    virtual R invoke(Args&&...args) && = 0;
  };
  template<class Sig,class F>
  struct pimpl;
  template<class R,class...Args,class F>
  struct pimpl<R(Args...),F>:ipimpl<R(Args...)> {
    F f;
    template<class Fin>
    pimpl(Fin&&fin):f(std::forward<Fin>(fin)){}
    R invoke(Args&&...args) && final override {
      return std::forward<F>(f)(std::forward<Args>(args)...);
    };
  };
  // void case,we don't care about what f returns:
  template<class...Args,class F>
  struct pimpl<void(Args...),F>:ipimpl<void(Args...)> {
    F f;
    template<class Fin>
    pimpl(Fin&&fin):f(std::forward<Fin>(fin)){}
    void invoke(Args&&...args) && final override {
      std::forward<F>(f)(std::forward<Args>(args)...);
    };
  };
}
template<class R,class...Args>
struct task_once<R(Args...)> {
  task_once(task_once&&)=default;
  task_once&operator=(task_once&&)=default;
  task_once()=default;
  explicit operator bool() const { return static_cast<bool>(pimpl); }

  R operator()(Args...args) && {
    auto tmp = std::move(pimpl);
    return std::move(*tmp).invoke(std::forward<Args>(args)...);
  }
  // if we can be called with the signature,use this:
  template<class F,class R2=R,std::enable_if_t<
        std::is_convertible<std::result_of_t<F&&(Args...)>,R2>{}
        && !std::is_same<R2,void>{}
    >* = nullptr
  >
  task_once(F&& f):task_once(std::forward<F>(f),std::is_convertible<F&,bool>{}) {}

  // the case where we are a void return type,we don't
  // care what the return type of F is,just that we can call it:
  template<class F,class=std::result_of_t<F&&(Args...)>,std::enable_if_t<std::is_same<R2,void>{}>* = nullptr
  >
  task_once(F&& f):task_once(std::forward<F>(f),bool>{}) {}

  // this helps with overload resolution in some cases:
  task_once( R(*pf)(Args...) ):task_once(pf,std::true_type{}) {}
  // = nullptr support:
  task_once( std::nullptr_t ):task_once() {}

private:
  std::unique_ptr< details_task_once::ipimpl<R(Args...)> > pimpl;

// build a pimpl from F.  All ctors get here,or to task() eventually:
  template<class F>
  task_once( F&& f,std::false_type /* needs a test?  No! */ ):
    pimpl( new details_task_once::pimpl<R(Args...),std::decay_t<F>>{ std::forward<F>(f) } )
  {}
  // cast incoming to bool,if it works,construct,otherwise
  // we should be empty:
  // move-constructs,because we need to run-time dispatch between two ctors.
  // if we pass the test,dispatch to task(?,false_type) (no test needed)
  // if we fail the test,dispatch to task() (empty task).
  template<class F>
  task_once( F&& f,std::true_type /* needs a test?  Yes! */ ):
    task_once( f?task_once( std::forward<F>(f),std::false_type{} ):task_once() )
  {}
};

live example.

请注意,您只能使用上述task_once在rvalue上下文中调用().这是因为()是破坏性的,因为它应该在你的情况下.

可惜上述依赖于C 14.而且我现在不喜欢写C11代码.所以,这是一个更简单的C 11解决方案,性能较差:

std::function<void()> a;
{
    Bar b;
    b.i = 10;
    auto pb = std::make_shared<Bar>(std::move(b));
    a = [pb]{ return foo(std::move(*pb)); };
}
a();

这将b的移动副本推送到共享指针中,将其存储在std ::函数中,然后在第一次调用()时破坏性地消耗它.

1它实现没有它的移动(除非它使用小功能优化,我希望它使用类型的移动).它还实现了转换为原始类型,但每种类型都支持.对于某些类型,它支持check-for-null(即,明确地转换为bool),但我真的不确定它所做的确切类型.

原文链接:https://www.f2er.com/c/113362.html

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