boost 绑定到数据成员回调行为
有人可以解释一下这段代码吗？
struct Class {
boost::function<void()> member;
};
Class c;
boost::function<boost::function<void()>()> foo = boost::bind(&Class::member, &c);
boost::function<void()> bar = boost::bind(&Class::member, &c);
为什么bar的定义能够编译，编译结果是什么？
编辑：foo() 按预期工作，调用 c.member()，但 bar() 却没有。
Can someone please explain this piece of code?
struct Class {
boost::function<void()> member;
};
Class c;
boost::function<boost::function<void()>()> foo = boost::bind(&Class::member, &c);
boost::function<void()> bar = boost::bind(&Class::member, &c);
Why does the definition of bar compile and what is the result of it?
Edit: foo() works as expected, calling c.member(), but bar() doesn't.
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第一个调用用于“生成”提取函子。该函子在被调用时将返回它所绑定的成员。
第二个调用只是隐藏传入的仿函数的返回类型（与第一个示例中的相同）。所以本质上，调用
bar不会执行任何操作。The first call is used to "generate" an extractor functor. That functor, when called, will return the member that it was bound to.
The second call just hides the return type of the functor that is passed in (which is the same as in the first example). So essentially, calling
barwill do nothing.如果您的类是这样的，您将需要绑定：
class Class {民众：
无效成员（）；
那么
你要做的是：
Class c;boost::function; the_function_i_want_to_call = boost::bind(&Class::member, c); the_function_i_want_to_call.call();You would need to bind if your class was like that:
class Class {public:
void member();
};
Then what you want to do is that :
Class c;boost::function<void()> the_function_i_want_to_call = boost::bind(&Class::member, c);the_function_i_want_to_call.call();