指向成员函数的指针问题

发布于 2024-09-29 03:59:19 字数 443 浏览 0 评论 0原文

在下面的代码中（请参阅评论）：

#include "stdafx.h"

#include <iostream>
using std::cout;

struct Base
{
 void fnc()
 {
  cout << "Base::fnc()";
 }

};

struct Impl
{
 void* data_;
 Impl(void (Base::*fp)())
 {
  fp();//HERE I'M INVOKING IT - I'M DOING SOMETHING WRONG!
 }
};
int _tmain(int argc, _TCHAR* argv[])
{
 return 0;
}

错误
“错误 1 错误 C2064：术语不计算为采用 0 个参数的函数” 为什么它不起作用以及如何修复它？

原文

In code below (please see comment):

#include "stdafx.h"

#include <iostream>
using std::cout;

struct Base
{
 void fnc()
 {
  cout << "Base::fnc()";
 }

};

struct Impl
{
 void* data_;
 Impl(void (Base::*fp)())
 {
  fp();//HERE I'M INVOKING IT - I'M DOING SOMETHING WRONG!
 }
};
int _tmain(int argc, _TCHAR* argv[])
{
 return 0;
}

Error
"Error 1 error C2064: term does not evaluate to a function taking 0 arguments"
Why doesn't it work and how to fix it?

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

汹涌人海 2024-10-06 03:59:19

它不起作用，因为 fp 不是函数指针而是成员指针。

如何修复它很简单，按照应该使用的方式使用它：someinstance.*fp();

回复收藏 0 原文

断爱 2024-10-06 03:59:19

问题是您将该函数作为自由函数调用，而实际上它不是。它是一个成员函数，您需要在对象的上下文中调用它：

(obj.*f)();

Boost.Bind 提供了一种惯用的方法来解决这个问题：

#include<boost/bind.hpp>

// ...
Impl i(boost::bind(&Base::fnc, obj));

您可以像这样定义 Impl 构造函数：

#include<boost/function.hpp>

// ...
Impl(boost::function<void ()> fnc)
{
    fnc();  // boost::bind translates it to obj.fnc()
}

如果只有 Impl 对象知道要调用该函数的对象，那么您可以使用 Boost.Bind 的占位符：

Impl i(boost::bind(&Base::fnc, boost::_1));

然后 Impl 构造函数将类似于

Impl(boost::function<void (Base)> fnc, Base& b)
{
    fnc(b);  // boost::bind translates it to b.fnc()
}

有时它是明智的使用模板接受函子的一侧：

template<class Op>
Impl(Op fnc) { ... }

因为这样客户端就可以传递任何成员函数，无论有没有 boost。但代价是您可能会遇到更难以理解的编译器错误消息。

The problem is that you are calling the function as a free function, when it isn't. It's a member function, and you need to call it in the context of an object:

(obj.*f)();

Boost.Bind offers an idiomatic way to tackle this:

#include<boost/bind.hpp>

// ...
Impl i(boost::bind(&Base::fnc, obj));

You can define the Impl constructor like so:

#include<boost/function.hpp>

// ...
Impl(boost::function<void ()> fnc)
{
    fnc();  // boost::bind translates it to obj.fnc()
}

If only the Impl object knows what object to call the function on, then you can use the placeholders of Boost.Bind:

Impl i(boost::bind(&Base::fnc, boost::_1));

And the Impl constructor would then be similar to

Impl(boost::function<void (Base)> fnc, Base& b)
{
    fnc(b);  // boost::bind translates it to b.fnc()
}

Sometimes it's wiser to use templates in the side that accepts the functor:

template<class Op>
Impl(Op fnc) { ... }

because then the client can pass any member function, with or without boost. But the cost is that you might have harder-to-understand compiler error messages.

回复收藏 0 原文

极致的悲 2024-10-06 03:59:19

typedef  int (MyClass::*memberPointer_t)(int);

...

memberPointer_t mb = &MyClass::function;
MyClass* object = getObject();

int returnValue = (object->*mb)(3);

...

由于它是指向成员函数的指针，因此必须在对象上调用它并使用 ->* 或 .* 运算符来调用它。

typedef  int (MyClass::*memberPointer_t)(int);

...

memberPointer_t mb = &MyClass::function;
MyClass* object = getObject();

int returnValue = (object->*mb)(3);

...

Since it's a pointer to a member function, you must call it on an object and use the ->* or the .* operator to call it.

回复收藏 0 原文