4

我无法理解为什么以下代码无法编译。

#include <memory>
#include <functional>

class Foo 
{
public:
  void Bar(int i) {}
};

void X(std::function<void(std::shared_ptr<Foo>)> f)
{

}

int main()
{
  std::shared_ptr<Foo> f(new Foo);
  auto f1(std::bind(&Foo::Bar, std::placeholders::_1, 1));
  X(f1);
  return 0;
}

g++ (4.6.3) 输出...

n file included from /usr/include/c++/4.6/memory:80:0,
                 from test.cpp:1:
/usr/include/c++/4.6/functional: In static member function ‘static void std::_Function_handler<void(_ArgTypes ...), _Functor>::_M_invoke(const std::_Any_data&, _ArgTypes ...) [with _Functor = std::_Bind<std::_Mem_fn<void (Foo::*)(int)>(std::_Placeholder<1>, int)>, _ArgTypes = {std::shared_ptr<Foo>}]’:
/usr/include/c++/4.6/functional:2148:6:   instantiated from ‘std::function<_Res(_ArgTypes ...)>::function(_Functor, typename std::enable_if<(! std::is_integral<_Functor>::value), std::function<_Res(_ArgTypes ...)>::_Useless>::type) [with _Functor = std::_Bind<std::_Mem_fn<void (Foo::*)(int)>(std::_Placeholder<1>, int)>, _Res = void, _ArgTypes = {std::shared_ptr<Foo>}, typename std::enable_if<(! std::is_integral<_Functor>::value), std::function<_Res(_ArgTypes ...)>::_Useless>::type = std::function<void(std::shared_ptr<Foo>)>::_Useless]’
test.cpp:19:7:   instantiated from here
/usr/include/c++/4.6/functional:1778:2: error: no match for call to ‘(std::_Bind<std::_Mem_fn<void (Foo::*)(int)>(std::_Placeholder<1>, int)>) (std::shared_ptr<Foo>)’
/usr/include/c++/4.6/functional:1130:11: note: candidates are:
/usr/include/c++/4.6/functional:1201:2: note: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) [with _Args = {_Args ...}, _Result = _Result, _Functor = std::_Mem_fn<void (Foo::*)(int)>, _Bound_args = {std::_Placeholder<1>, int}]
/usr/include/c++/4.6/functional:1215:2: note: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const [with _Args = {_Args ...}, _Result = _Result, _Functor = std::_Mem_fn<void (Foo::*)(int)>, _Bound_args = {std::_Placeholder<1>, int}]
/usr/include/c++/4.6/functional:1229:2: note: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) volatile [with _Args = {_Args ...}, _Result = _Result, _Functor = std::_Mem_fn<void (Foo::*)(int)>, _Bound_args = {std::_Placeholder<1>, int}]
/usr/include/c++/4.6/functional:1243:2: note: template<class ... _Args, class _Result> _Result std::_Bind<_Functor(_Bound_args ...)>::operator()(_Args&& ...) const volatile [with _Args = {_Args ...}, _Result = _Result, _Functor = std::_Mem_fn<void (Foo::*)(int)>, _Bound_args = {std::_Placeholder<1>, int}]
4

1 回答 1

5

这是我在 4.7 中修复的 GCC 错误，但我不记得确切是哪个错误。我会试着弄清楚...

编辑：啊哈，它是PR 55463，所以在 4.7 中没有固定，它只在 GCC 主干上固定（什么是 4.8）

错误是返回的调用包装器mem_fn不接受右值，并且您的std::function<void(std::shared_ptr<Foo>)类型将右值传递shared_ptr<Foo>给调用包装器返回的bind.

作为一种解决方法，您可以将函数签名更改为：

void X(std::function<void(const std::shared_ptr<Foo>&)> f)

我不认为我可以将修复程序反向移植到 4.7 分支，因为由于那个错误，我对它进行了一些相当大的更改，mem_fn这并不适合稳定的发布分支（我还在标准中发现了一个新缺陷)

于 2012-12-19T19:16:50.477 回答