Richter substitution principle (try not to rewrite the parent class method by adding the base class to let the atomic class and the original parent class reduce coupling through aggregation, composition, and dependency)

basic concepts:

1. The Richter substitution principle was proposed in 1988 by a girl from the Massachusetts Institute of Technology.
2. When inheriting, try not to override the parent class methods in the subclass
    (for example: all the methods of class A are overridden by class B. Then why inherit? Just create a new class B directly)
3. The Richter substitution principle tells us that inheritance enhances the coupling of two classes , and solves problems through aggregation, composition, and dependency under appropriate circumstances.

What to do with Richter replacement:

      Increase base class, reduce coupling

For example:

public class Liskov {

	public static void main(String[] args) {
		A a = new A();
		System.out.println("11-3=" + a.func1(11, 3));
		System.out.println("1-8=" + a.func1(1, 8));

		System.out.println("-----------------------");
		B b = new B();
		System.out.println("11-3=" + b.func1(11, 3));//这里本意是求出11-3
		System.out.println("1-8=" + b.func1(1, 8));// 1-8
		System.out.println("11+3+9=" + b.func2(11, 3));
		
	}

}
// A类
class A {
	// 返回两个数的差
	public int func1(int num1, int num2) {
		return num1 - num2;
	}
}

Pay attention to the fun1 method here. Programmer 1 accidentally re-parented the method, and the name may happen to be the same, resulting in the 11-3=14,1-9=9 requested above. 

// B类继承了A
// 增加了一个新功能：完成两个数相加,然后和9求和
class B extends A {
	//这里，重写了A类的方法, 可能是无意识
	public int func1(int a, int b) {
		return a + b;
	}

	public int func2(int a, int b) {
		return func1(a, b) + 9;
	}
}

 

Solution:

public class Liskov {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		A a = new A();
		System.out.println("11-3=" + a.func1(11, 3));
		System.out.println("1-8=" + a.func1(1, 8));

		System.out.println("-----------------------");
		B b = new B();
		//因为B类不再继承A类，因此调用者，不会再func1是求减法
		//调用完成的功能就会很明确
		System.out.println("11+3=" + b.func1(11, 3));//这里本意是求出11+3
		System.out.println("1+8=" + b.func1(1, 8));// 1+8
		System.out.println("11+3+9=" + b.func2(11, 3));
		
		
		//使用组合仍然可以使用到A类相关方法
		System.out.println("11-3=" + b.func3(11, 3));// 这里本意是求出11-3
		

	}

}

//创建一个更加基础的基类
class Base {
	//把更加基础的方法和成员写到Base类
}

// A类
class A extends Base {
	// 返回两个数的差
	public int func1(int num1, int num2) {
		return num1 - num2;
	}
}

// B类继承了A
// 增加了一个新功能：完成两个数相加,然后和9求和
class B extends Base {
	//如果B需要使用A类的方法,使用组合关系
	private A a = new A();
	
	//这里，重写了A类的方法, 可能是无意识
	public int func1(int a, int b) {
		return a + b;
	}

	public int func2(int a, int b) {
		return func1(a, b) + 9;
	}
	
	//我们仍然想使用A的方法
	public int func3(int a, int b) {
		return this.a.func1(a, b);
	}
}