遇到 error LNK2019: 无法解析的外部符号

创建自己定义的Screen类，在定义函数时，将成员函数定义为inline函数，链接时报出error：

>demo.obj : error LNK2019: 无法解析的外部符号
> "public: class Screen & __thiscall Screen::set(char)" (?set@Screen@@QAEAAV1@D@Z)
> 该符号在函数 _main 中被引用

删除定义处的inline标识就没问题了，很奇怪为什么。
Screen.h

#pragma once

#include 
#include 

class Screen {
public:
   typedef std::string::size_type pos;
   Screen() = default;

   Screen(pos ht, pos wd)
      : content(ht * wd, ' ') {}

   Screen(pos ht, pos wd, char c)
      : height(ht), width(wd), content(ht * wd, c) {};

   char get() const {
      return content[cursor];
   }

   char get(pos ht, pos wd) const;

   Screen &move(pos r, pos c);

   void accessTime() const {
      ++times;
   }

   Screen &set(char);
   Screen &set(pos, pos, char);

   const Screen &display(std::ostream &os) const {
      os << content;
      return *this;
   }

   Screen &display(std::ostream &os) {
      os << content;
      return *this;
   }

   ~Screen() {};

private:
   void do_display(std::ostream &os) const {
      os << content;
   }

   pos cursor = 0;
   pos height = 0, width = 0;
   std::string content;

   mutable size_t times = 0;
};

Screen.cpp

#include "pch.h"
#include "Screen.h"


char Screen::get(pos r, pos c) const {
   pos row = r * width;
   return content[row + c];
}

Screen &Screen::move(pos r, pos c) {
   pos row = r * width;
   cursor = row + c;
   return *this;
}

inline Screen& Screen::set(char c) {
   content[cursor] = c;
   return *this;
}

Screen& Screen::set(pos ht, pos wd, char c) {
   content[ht * width + wd] = c;
   return *this;
}

main.cpp

Screen myScreen(5, 5, 'X');
   myScreen.move(4, 0).set('#').display(cout);
   cout << "\n";
   myScreen.display(cout);
   cout << "\n";

常见的该问题的解决方法可以参考：https://blog.csdn.net/qq_25810695/article/details/80282515