搜索一个优雅和非侵入性的方式来访问类的私有方法 [英] Search for an elegant and nonintrusive way to access private methods of a class

问题描述

免责声明：这绝不意味着在生产代码中使用。这是一个在C ++边缘的探索：）

Disclaimer: This is never meant to be used in production code. It's an exploration at the edges of C++ :)

我的问题是一个跟进，基于与@Johannes Schaub的讨论：
在c ++中调用私有方法。

My question is a follow up, based on a discussion with @Johannes Schaub here: calling private methods in c++.

I在他的博客中找到了一个非常短的私人成员访问解决方案：
http://bloglitb.blogspot.de/2011/12/access-to-private-members-safer.html

I found a very short solution for private member access on his blog: http://bloglitb.blogspot.de/2011/12/access-to-private-members-safer.html

这里是示例：

#include <iostream>
using namespace std;

// example class
struct A {
  A(int a, double b):_a(a),_b(b) { }
private:
  int _a;
  double _b;
  int f() { return _a; }
public:
};

//Robber template: provides a legal way to access a member
template<typename Tag, typename Tag::type M>
struct Rob { 
  friend typename Tag::type get(Tag) {
    return M;
  }
};
// tag used to access A::_a
struct A_access_a 
{ 
  typedef int A::*type;
  friend type get(A_access_a);
};

// Explicit instantiation; the only place where it is legal to pass the address of a private member.
template struct Rob<A_access_a, &A::_a>;

int main() {

    A sut(42, 2.2);

    int a = sut.*get(A_access_a());
    cout << a << endl;

    return 0;
}

我想知道这种非常优雅的方法是否可以重用来从外部访问私有方法

I wondered if this very elegant approach can be reused to access private methods from outside of a class.

我想要的是一个方法调用的简单方法：

What I would like to have, is the same simple approach for a method call:

struct A_access_f
{
    typedef int (A::*type)();
    friend type get(A_access_f);
};
template struct Rob<A_access_f, &A::f>;

是否可以运行？

这是我现在最好的尝试：

This is my best attempt till now:

typedef int (A::*pf)();
pf func = sut.*get(A_access_f());

我的编译器仍然抱怨：

prog.cpp：45：33：错误：无效使用非静态成员函数
pf func = sut。* get（A_access_f（））;

prog.cpp:45:33: error: invalid use of non-static member function pf func = sut.*get(A_access_f());

推荐答案

你几乎在那里。这里是你应该写的：

You were almost there. Here is what you should have written:

typedef int (A::*pf)();
const pf func = get(A_access_f());
int a = (sut.*func)();

或作为（难以摘要）单行：

Or as a (hard-to-digest) one-liner:

int a = (sut.*get(A_access_f()))();

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