循环依赖与前向声明错误

问题描述

在A.hpp文件中，我有一个结构，其中有一个B类的指针

  struct state 
 {
 B * b; 
}; 
  

在A.hpp文件中，我添加了一个forward声明， cpp文件

  // A.hpp 
类B 
  pre> 
 
 
在B.hpp文件中，函数使用在A.hpp中声明的状态作为函数的参数。
  bool function_in_b（state * s）
  
我还在B.hpp文件中添加了A的前向声明，并在B.cpp文件中添加了A，A.hpp的头文件。
  // B.hpp 
 class A 
  
头护板。 
如果我尝试编译，它不会找到'状态'声明在A.hpp。 
因此，它不会找到匹配的函数，并抱怨候选人
  bool function_in_b b $ b  
如何解决这个问题？
 
 
 
提前感谢
解决方案
在 B.hpp  -declared  A ，但不是 state   - 所以当它第一次看到 function_in_b ）它不知道状态是什么。当您在 B.cpp 中包含 A.hpp 时，已经太晚了。您需要在 B.hpp 中转发声明 state ，即
  struct state; 
 
 bool function_in_b（state * s） 
  
 
In A.hpp file I have a structure, which has a pointer of B class
struct state
{
    B  *b;
};
In A.hpp file, I added a forward declaration and I included B.hpp file in A.cpp file
//A.hpp
class B
In B.hpp file, a function uses the state, which declared in A.hpp as an argument on the function.
bool function_in_b(state *s)
I also added a forward declaration of A in B.hpp file and I added the header file of A, A.hpp in B.cpp file.
//B.hpp
class A
All header files have a header guard.If I try to compile, it won't find 'state' declared in A.hpp.Thus, it won't find the matching function and complains the candidates are
bool function_in_b(int *)
How do I fix this problem?
Thanks in advance
 解决方案 
In B.hpp, you say you forward-declared A, but not state - so when it first sees function_in_b(state *s) it doesn't know what state is. By the time you include A.hpp in B.cpp it's too late. You need to forward declare state in B.hpp, i.e.
struct state;

bool function_in_b(state *s);
                        
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