为什么在使用boost :: function调用函数对象时不需要类实例

Question

#include <iostream>
#include <vector>
#include <string>
#include <ostream>
#include <algorithm>

#include <boost/function.hpp>
using namespace std;

class some_class
{
public:
  void do_stuff(int i) const
  {
    cout << "some_class i: " << i << endl;
  }
};

class other_class
{
public:
  void operator()(int i) const
  {
    cout << "other_class i: " << i << endl;
  }
};

int main() {
  //             CASE ONE
  boost::function<void (some_class, int) > f;
  // initilize f with a member function of some_class
  f = &some_class::do_stuff;
  // pass an instance of some_class in order to access class member
  f(some_class(), 5); 

  //             CASE TWO
  boost::function<void (int) > f2;
  // initialize f2 with a function object of other_class
  f2 = other_class();
  // Note: directly call the operator member function without
  // providing an instance of other_class
  f2(10);
}


// output
~/Documents/C++/boost $ ./p327
some_class i: 5
other_class i: 10

问题 >当我们通过boost :: function调用一个函数对象时,为什么我们不必为该类提供一个实例来调用这个类成员函数？

是因为我们通过以下方式提供了这样的信息吗？

f2 = other_class();

Answer 1

您必须为该类提供一个实例,并且您正在提供一个实例.

boost::function<void (int) > f2;
f2 = other_class();

这构造了一个other_class对象,并将该对象分配给f2.boost::function然后复制该对象,以便在您尝试调用它时,您不需要再次实例化它.