关于public String checkUser(User one, User two)

我把public String checkUser(User one, User two)这句中的String改成void了才运行出，麻烦老师帮忙看看，哪些地方改了才能正确运行呢？

public class User {

// 定义属性用户名、密码

private String name;

private String password;

User(String name,String password){

this.name=name;

this.password=password;

}

public String getName() {

return "用户名："+name;

}

public void setName(String name) {

this.name = name;

}

public String getPassword() {

return "密码："+password;

}

public void setPassword(String password) {

this.password = password;

}

}

=====================================================

public class UserManager {

// 用户信息验证的方法

    public void checkUser(User one, User two) {

if(one.equals(two.getName())&&one.equals(two.getPassword())) {

System.out.println("用户名和密码一致");

}else {

System.out.println("用户名或密码不一致");

}

       // 判断用户名和密码是否为空，是否一致

    }

}

=================================================

public class UserTest {

public static void main(String[] args) {

//实例化对象，调用相关方法实现运行效果

User user1=new User("Lucy","123456");

User user2=new User("Mike","123456");

System.out.println(user1.getName());

System.out.println(user1.getPassword());

System.out.println(user2.getName());

System.out.println(user2.getPassword());

System.out.println("=========================");

UserManager user=new UserManager();

user.checkUser(user1, user2);

}

}