我试图创建一个类,在构造函数中将一个lambda函数作为参数,我希望这个函数成为类的朋友.该类的代码如下所示：

using func = std::function<void(void)>;    

class foo
{
public:
    foo(func f)
    {
        this->f = f;
    }

    func f;
private:
    int value_I_want_to_modify; //an int I want to change from the function I've passed in the constructor
}

在main()中我会写这样的东西：

int main()
{
    //this will give an error because I cannot access private members from outside class
    foo v
    {
        [&v](void) { v.value_I_want_to_modify = 0 };
    }
}

现在我希望函数成为该类的朋友,但我找不到办法.

解决方法:

How to make a lambda function friend of a class?

你不能.这是一个问题22.

如果在定义类之前定义lambda函数,则无法访问类的成员变量.

using func = std::function<void(void)>;    

class foo;

// Trying to define the lambda function before the class.
// Can't use f.value_I_want_to_modify since foo is not defined yet.
auto lambda_function = [](foo& f) { f.value_I_want_to_modify = 0;}

class foo
{
   public:
      foo(func f)
      {
         this->f = f;
      }

      func f;
   private:
      int value_I_want_to_modify;
};

int main()
{
    foo v{lambda_function};
}

如果在定义类之后定义lambda函数,则不能使lambda函数成为类的朋友.

using func = std::function<void(void)>;

class foo
{
   public:
      foo(func f)
      {
         this->f = f;
      }

      func f;
   private:
      int value_I_want_to_modify;
};

int main()
{
   foo f
   {
      // Can't make the lambda function a friend of foo
      // since it cannot be declared before the class definition.
      [&f](void) { f.value_I_want_to_modify = 0;}
   }
}

最简单的解决方法是修改lambda函数以接受int&作为参数并修改其值.

#include <functional>

using func = std::function<void(int&)>;

class foo
{
   public:
      foo(func f)
      {
         this->f = f;
         this->f(value_I_want_to_modify);
      }

   private:

      func f;
      int value_I_want_to_modify;
};

int main()
{
   foo v{ [](int& out) { out = 0;} };
}