如何在这个类中使用函数指针？ [英] How do I use function pointers within this class?

问题描述

我有一个类：Circle。它可以绘制一个实心圆或一个轮廓的圆。基于它的当前设置，我想要一个draw（）方法调用draw_filled（）或draw_outlined（）。

在我的类中，我有成员 void（* draw）（void）;

我有两个私有函数： void draw_filled ）; void draw_outlined（）;

然后，我有以下方法：

  void Circle :: fill（const bool fill）
 {
 m_fill = fill; 
 
 if（fill）
 draw = draw_filled; 
 else 
 draw = draw_outlined; 
} 
  

我在两个绘图赋值中都遇到错误：

错误C2440：'='：无法从'void（__thiscall X2D :: GL :: Circle :: *）（void）'转换为'void（__cdecl *） $ b 1>没有上下文可以进行这种转换

通常情况下，我不在类中使用函数指针，所以这对我来说是新的。任何帮助将不胜感激。谢谢。

解决方案

从错误消息中可以看出 draw 是指向函数的通用指针，而不是指向成员函数的指针。 draw的声明应该是 void（X2D :: GL :: Circle :: * draw）（void），这样 draw 指向 Circle 的成员函数。

I have a class: Circle. It can draw either a filled circle or an outlined circle. Based on its current setting, I want a general draw() method that calls either draw_filled() or draw_outlined().

In my class, I have the member void (*draw)(void);

I have two private functions: void draw_filled(); void draw_outlined();

Then, I have the following method:

void Circle::fill(const bool fill)
{
    m_fill = fill;

    if (fill)
        draw = draw_filled;
    else
        draw = draw_outlined;
}

I get an error on both draw assignments:

error C2440: '=' : cannot convert from 'void (__thiscall X2D::GL::Circle::* )(void)' to 'void (__cdecl *)(void)' 1> There is no context in which this conversion is possible

Normally, I don't use function pointers in classes, so this is new to me. Any help would be appreciated. Thank you.

解决方案

From the error message, it appears that draw is a generic pointer to a function, and not a pointer to a member function. The declaration of draw should be void (X2D::GL::Circle::*draw)(void), so that draw points to a member function of Circle.

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