Question

在Java中，如何根据其构造函数选择其函数的Runnable类？

例如：

Class A implements Runnable {

    public A(int var) {
        //things go here
    }

    public A(int var1, String var2) {
        //things go here
    }

    public void run() {
        //do (a) if Constructor A(int var)
        //do (b) if Constructor A(int var1, String var2)
    }
}

Answer 1

你可以这样做：

class A implements Runnable {
  int constructor;

public A(int var) {
    //things go here
    constructor = 1;
}

public A(int var1, String var2) {
    //things go here
    constructor = 2; 
}

public void run() {
    if (constructor == 1) ...
    if (Constructor == 2) ...
}

}

但我建议使用多态。看起来这是两类不同的对象。

Answer 2

我会设置一个Strategy模式：

class A implements Runnable {

    private final ExecutionStrategy strategy;

    public A(int var) {   
        strategy = new FooStrategy();
    }

    public A(int var1, String var2) {
        strategy = new BarStrategy();
    }

    public void run() {
        strategy.exec();
    }

    interface ExecutionStrategy { 
        public void exec();
    }

    private class FooStrategy implements ExecutionStrategy { 
        @Override
        public void exec() { 
        //implement me
        }
    }

    private class BarStrategy implements ExecutionStrategy { 
        @Override
        public void exec() { 
        //implement me
        }
    }
}

Runnable函数取决于构造函数

2 个答案: