c++ - “类形状”没有名为“信息”的成员，但添加一个也不起作用

Question

我正在尝试编译一些代码（此代码）但是当我注释掉第 25 行时：
virtual void info()=0;
它无法编译：

shape.cpp: In function ‘int main()’:
shape.cpp:345:11: error: ‘class shape’ has no member named ‘info’
  svec[0]->info();

但是保留第 25 行会给纯虚函数带来很长的错误...

shape.cpp:77:15: error: cannot declare parameter ‘squ’ to be of abstract type ‘square’
   cube(square squ):
               ^
shape.cpp:30:7: note:   because the following virtual functions are pure within ‘square’:
 class square : public shape {
       ^
shape.cpp:25:16: note:  virtual void shape::info()
   virtual void info()=0;
                ^
shape.cpp:167:20: error: cannot declare parameter ‘rec’ to be of abstract type ‘rectangle’
   cuboid(rectangle rec, double d):
                    ^
shape.cpp:110:7: note:   because the following virtual functions are pure within ‘rectangle’:
 class rectangle : public shape {
       ^
shape.cpp:25:16: note:  virtual void shape::info()
   virtual void info()=0;

等等...

谁能告诉我我做错了什么？谢谢。

Answer 1

该函数const在派生类中声明，但不在基类中声明。这意味着派生类不会覆盖该函数；他们只是声明了一个同名的不同函数。

要么添加const基类，要么在派生类中删除它。