declare the definition of a class separating from its body

Following code is in the same header file

#include <iostream>
               
class B
{
public:
    void doSth()
    {  
        A::func(1);
    } 
};

class A
{
public:
    static void func(int a)
    {
        std::cout<<"a = "<< a <<std::endl;
    }
};

If class A is defined after class B like above, the compiler will complain about ‘func’ is not a member 'A'.

My question is, is there anyway to solve this issue without changing the order how class B and class A are defined? Like putting a declaration of class A somewhere above class B? Thanks.

Solution

If you want B::doSth() to be declared and defined completely inside of B's class declaration, then you must move the full declaration of A ahead of B, eg:

#include <iostream>
               
class A
{
public:
    static void func(int a)
    {
        std::cout << "a = " << a << std::endl;
    }
};  

class B
{
public:
    void doSth()
    {  
        A::func(1);
    } 
};

Otherwise, if you don't need B::doSth() to be completely inside the B class declaration then you can separate its declaration and definition, and then move the definition after the declaration of the A class, eg:

#include <iostream>
               
class B
{
public:
    void doSth();
};

class A
{
public:
    static void func(int a)
    {
        std::cout << "a = " << a << std::endl;
    }
};

inline void B::doSth()
{  
    A::func(1);
}

Preferably, method definitions that are not inside their class declaration should be implemented in separate .cpp files instead, eg:

A.h

#ifndef A_H
#define A_H

class A
{
public:
    static void func(int a);
};

#endif

A.cpp

#include <iostream>               
#include "A.h"

void A::func(int a)
{
    std::cout << "a = " << a << std::endl;
}

B.h

#ifndef B_H
#define B_H

class B
{
public:
    void doSth();
};

#endif

B.cpp

#include "B.h"
#include "A.h"
               
void B::doSth()
{  
    A::func(1);
}