传递和调用成员函数（boost :: bind / boost :: function？） [英] Pass and call a member function (boost::bind / boost::function?)

问题描述

我可能有一个很尴尬的简单问题：在类中传递和调用成员函数。我知道我想使用BOOST绑定（和或函数），但我还没有真正把握它的概念。

下面的代码编译和执行与问题。但是当我想将f3函数改为非静态类函数时，有趣的开始：

  #include < iostream> 
 #include< inttypes.h> 
 #include< boost / bind.hpp> 
 #include< boost / function.hpp> 
 
 class Test 
 {
 public：
 void f1（）; 
 private：
 void f2（void（* callfunc）（uint32_t））; 
 static void f3（uint32_t x）; 
}; 
 
 void Test :: f1（）{
 f2（f3）; 
} 
 
 void Test :: f2（void（* callfunc）（uint32_t））{
（* callfunc）（42）; 
} 
 
 void Test :: f3（uint32_t x）{
 std :: cout< x：<< x<< std :: endl; 
} 
 
 int main（int argc，char ** argv）
 {
 Test ct; 
 ct.f1（）; 
 return 0; 
} 
  

现在，更改后

  static void f3（uint32_t x）; 
  

到

  void f3（uint32_t x）; 
  

编译器不开心，告诉我错误：没有匹配函数调用' ：f2（）'

通过阅读了很多关于boost :: bind和boost :: function的SO文章，我想我需要改变f2的定义（）和如何f1（）调用f2（）给f3（）作为目标调用，但除此之外...关于boost :: bind和boost函数的每一个组合我试图惨败编译失败。

我该如何写这篇文章？作为一个额外的问题：是否有任何简单的介绍读取boost :: bind和boost :: function？

解决方案

p> boost :: function是一个模板类，它接受一个函数签名。您也可以使用function0，function1等。

boost :: function< void（uint32_t）>

定义一个看起来像一个函数的可调用，即它需要一个类型 uint32_t 并返回void。

相应的编号模板是 function1< void，uint32_t> 。这些总是首先指示返回类型，然后按顺序显示参数。

boost :: bind 函数，它推导你传递给它的参数，并为你创建一个函数。

它不会为你创建一个void（uint32_t）

因此将您的签名更改为：

  f2（boost :: function< void（uint32_t）>）; 
  

然后你可以这样调用：

  f2（boost :: bind（& Test :: f3，this，_1））; 
 

请注意，奇怪的_1是一个占位符告诉boost :: bind它需要放在参数，在这种情况下为 uint32_t

I have a probably embarassingly simple problem: pass and call a member function in a class. I know I want to use BOOST bind (and or function), but I haven't really grasped the concept to it yet.

The following code compiles and executes with problem. But when I want to change the "f3" function to a non-static class function, then the fun begins:

#include <iostream>
#include <inttypes.h> 
#include <boost/bind.hpp>
#include <boost/function.hpp>

class Test
{
public:
  void f1();
private:
  void f2(void (*callfunc)(uint32_t));
  static void f3(uint32_t x);
};

void Test::f1(){
  f2(f3);
}

void Test::f2(void (*callfunc)(uint32_t)){
  (*callfunc)(42);
}

void Test::f3(uint32_t x){
  std::cout << "x: " << x << std::endl;
}

int main(int argc, char ** argv)
{
  Test ct;
  ct.f1();
  return 0;
}

Now, after changing

static void f3(uint32_t x);

to

void f3(uint32_t x);

the compiler isn't happy and tells me "error: no matching function for call to 'Test::f2()'"

Having read through a number of SO posts regarding boost::bind and boost::function, I think I need to change the definition of f2() and how f1() calls f2() giving f3() as target to call, but apart from that ... about every combination of boost::bind and boost function I tried miserably fails to compile.

How do I need to write this? As a bonus question: are there any simple introductory reads on boost::bind and boost::function? The BOOST docs did not really help me there.

B.

解决方案

boost::function is a template class, that takes a function signature. You can also use function0, function1, etc.

boost::function< void(uint32_t) >

defines a "callable" that looks like a function, i.e. it takes a single parameter of type uint32_t and returns void.

The appropriate numbered template is function1< void, uint32_t >. These always indicate the return type first, then the parameters in order.

boost::bind is a very special function that deduces the arguments you pass into it and creates a functor for you.

It will not create a void(uint32_t) for you, it will create something that has the pattern of one.

Therefore change your signature to:

void f2(boost::function<void(uint32_t)>);

Then you can call it like this:

f2( boost::bind( &Test::f3, this, _1 ) );

Note the strange _1 is a "placeholder" telling boost::bind where it needs to put in the parameter, in this case the uint32_t

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