Question

我只是写一个简单的例子来测试boost :: bind。我用它来实例化模板成员函数，但它不能用g ++ 4.6.0编译。我不知道是什么问题。这是代码：

#include <boost/bind.hpp>

struct Functor
{
  void operator()()
  {

  }
};

struct DerivedFinishAction
{
  DerivedFinishAction()
  {}

  void Inc()
  {

  }

  template <typename T>
  void TmplFunc(T t)
  {
    (boost::bind(&DerivedFinishAction::BindFunc<T>, this , t))();
  }

  template <typename T>
  void BindFunc(T t)
  {
    t();
  }

  void Func()
  {
    Functor f;
    TmplFunc(f); // this is OK
    TmplFunc(boost::bind(&DerivedFinishAction::Inc, this)); // compile error
  }
};

int main(int argc, char *argv[])
{

  return 0;
}

g ++给出了以下错误：

In file included from /usr/include/boost/bind.hpp:22:0,
                 from testBind.cpp:1:
/usr/include/boost/bind/bind.hpp: In member function ‘void boost::_bi::list2<A1, A2>::operator()(boost::_bi::type<void>, F&, A&, int) [with F = boost::_mfi::mf1<void, DerivedFinishAction, boost::_bi::bind_t<void, boost::_mfi::mf0<void, DerivedFinishAction>, boost::_bi::list1<boost::_bi::value<DerivedFinishAction*> > > >, A = boost::_bi::list0, A1 = boost::_bi::value<DerivedFinishAction*>, A2 = boost::_bi::bind_t<void, boost::_mfi::mf0<void, DerivedFinishAction>, boost::_bi::list1<boost::_bi::value<DerivedFinishAction*> > >]’:
/usr/include/boost/bind/bind_template.hpp:20:59:   instantiated from ‘boost::_bi::bind_t<R, F, L>::result_type boost::_bi::bind_t<R, F, L>::operator()() [with R = void, F = boost::_mfi::mf1<void, DerivedFinishAction, boost::_bi::bind_t<void, boost::_mfi::mf0<void, DerivedFinishAction>, boost::_bi::list1<boost::_bi::value<DerivedFinishAction*> > > >, L = boost::_bi::list2<boost::_bi::value<DerivedFinishAction*>, boost::_bi::bind_t<void, boost::_mfi::mf0<void, DerivedFinishAction>, boost::_bi::list1<boost::_bi::value<DerivedFinishAction*> > > >, boost::_bi::bind_t<R, F, L>::result_type = void]’
testBind.cpp:24:5:   instantiated from ‘void DerivedFinishAction::TmplFunc(T) [with T = boost::_bi::bind_t<void, boost::_mfi::mf0<void, DerivedFinishAction>, boost::_bi::list1<boost::_bi::value<DerivedFinishAction*> > >]’
testBind.cpp:37:58:   instantiated from here
/usr/include/boost/bind/bind.hpp:313:9: error: invalid use of void expression

有人可以帮忙解释一下吗？为什么第一个实例化是正常的，而第二个实例化导致编译错误？

Answer 1

此处涉及boost::bind的（非显而易见的）功能。 http://www.boost.org/libs/bind/#nested_binds

如果你写：

void func1(int len) {return len+1;};
int func2(std::string str) {return str.length();};

assert(
    boost::bind(func1, boost::bind(func2, _1) )("Hello")
    == 6 );

boost::bind假设您的意思是“在func2上运行"Hello"，然后在结果上运行func1”。这允许更有趣的部分功能应用。

在你的程序中，你的表达式相当于：

boost::bind(&DerivedFinishAction::BindFunc<...>, 
            this, 
            boost::bind(&DerivedFinishAction::Inc, this))

因此boost::bind会尝试弄清楚如何在其参数上运行DerivedFinishAction::Inc，因此它可以将该结果传递给DerivedFinishAction::BindFunc<...>。但DerivedFinishAction::Inc返回void，无法传递给DerivedFinishAction::BindFunc<...>。因此，您会收到编译器错误：

/usr/include/boost/bind/bind.hpp:313:9: error: invalid use of void expression

修改：根据文档，您可以使用protect来实现所需的行为：

#include <boost/bind/protect.hpp>
...
TmplFunc(boost::protect(boost::bind(&DerivedFinishedAction::Inc, this))); // no longer an error
...

boost :: bind不使用成员模板函数进行编译

1 个答案: