流水线设计模式实现

Question

public interface Step<T, U> {
    public U execute(T input);
}

public class StepOne implements Step<Integer, Integer> {
    @Override
    public Integer execute(Integer input) {
        return input + 100;
    }
}

public class StepTwo implements Step<Integer, Integer> {
    @Override
    public Integer execute(Integer input) {
        return input + 500;
    }
}

public class StepThree implements Step<Integer, String> {
    @Override
    public String execute(Integer input) {
        return "The final amount is " + input;
    }
}

public class Pipeline {
    private List<Step> pipelineSteps = new ArrayList<>();
    private Object firstStepInput = 100;

    public void addStep(Step step) {
        pipelineSteps.add(step);
    }

    public void execute() {
        for (Step step : pipelineSteps) {
            Object out = step.execute(firstStepInput);
            firstStepInput = out;
        }
   }
}

public class Main {
    public static void main(String[] args) {
        Pipeline pipeline = new Pipeline();
        pipeline.addStep(new StepOne());
        pipeline.addStep(new StepTwo());
        pipeline.addStep(new StepThree());

        pipeline.execute();
    } 
}

有人可以通过添加到我已经编码的内容来帮助我设计解决方案吗？或者向我指出一个已经存在的模式来解决这个问题？

Answer 1

为什么需要额外的管道类？我想你可以除掉中间人。这将使您的api更加简单，例如：

Step<Integer, String> source = Step.of(Object::toString);
Step<Integer, Integer> toHex = source.pipe(it -> Integer.parseInt(it, 16));

toHex.execute(11/*0x11*/);// return 17;

您可以简单地在java-8中实现管道模式，如下所示：

interface Step<I, O> {

    O execute(I value);

    default <R> Step<I, R> pipe(Step<O, R> source) {
        return value -> source.execute(execute(value));
    }

    static <I, O> Step<I, O> of(Step<I, O> source) {
        return source;
    }
}

在以前的java版本中，您可以使用抽象类：

abstract static class Step<I, O> {

    public abstract O execute(I value);

    public <R> Step<I, R> pipe(Step<O, R> source) {
        return new Step<I, R>() {
            @Override
            public R execute(I value) {
                return source.execute(Step.this.execute(value));
            }
        };
    }

    public static <I, O> Step<I, O> of(Step<I, O> source) {
        return source;
    }
}