代码重构

1.如何去除多余的if else

• 出现if/else和switch/case的场景

通常业务代码会包含这样的逻辑：每种条件下会有不同的处理逻辑。比如两个数a和b之间可以通过不同的操作符（+，-，*，/）进行计算，初学者通常会这么写：

public int calculate(int a, int b, String operator) {
    int result = Integer.MIN_VALUE;
 
    if ("add".equals(operator)) {
        result = a + b;
    } else if ("multiply".equals(operator)) {
        result = a * b;
    } else if ("divide".equals(operator)) {
        result = a / b;
    } else if ("subtract".equals(operator)) {
        result = a - b;
    }
    return result;
}

这种最基础的代码如何重构呢？

• 工厂类

public class OperatorFactory {
    static Map<String, Operation> operationMap = new HashMap<>();
    static {
        operationMap.put("add", new Addition());
        operationMap.put("divide", new Division());
        // more operators
    }
 
    public static Optional<Operation> getOperation(String operator) {
        return Optional.ofNullable(operationMap.get(operator));
    }
}

• 枚举

public enum Operator {
    ADD {
        @Override
        public int apply(int a, int b) {
            return a + b;
        }
    },
    // other operators
    
    public abstract int apply(int a, int b);

}

• Command模式

public class AddCommand implements Command {
    // Instance variables
 
    public AddCommand(int a, int b) {
        this.a = a;
        this.b = b;
    }
 
    @Override
    public Integer execute() {
        return a + b;
    }
}

• 规则引擎

1、定义规则

public interface Rule {
    boolean evaluate(Expression expression);
    Result getResult();
}

2、Add 规则

public class AddRule implements Rule {
    @Override
    public boolean evaluate(Expression expression) {
        boolean evalResult = false;
        if (expression.getOperator() == Operator.ADD) {
            this.result = expression.getX() + expression.getY();
            evalResult = true;
        }
        return evalResult;
    }    
}

3、表达式

public class Expression {
    private Integer x;
    private Integer y;
    private Operator operator;        
}

4、规则引擎

public class RuleEngine {
    private static List<Rule> rules = new ArrayList<>();
 
    static {
        rules.add(new AddRule());
    }
 
    public Result process(Expression expression) {
        Rule rule = rules
          .stream()
          .filter(r -> r.evaluate(expression))
          .findFirst()
          .orElseThrow(() -> new IllegalArgumentException("Expression does not matches any Rule"));
        return rule.getResult();
    }
}

• 策略模式

1、操作

public interface Opt {
    int apply(int a, int b);
}

@Component(value = "addOpt")
public class AddOpt implements Opt {
    @Autowired
    xxxAddResource resource; // 这里通过Spring框架注入了资源

    @Override
    public int apply(int a, int b) {
       return resource.process(a, b);
    }
}

@Component(value = "devideOpt")
public class devideOpt implements Opt {
    @Autowired
    xxxDivResource resource; // 这里通过Spring框架注入了资源

    @Override
    public int apply(int a, int b) {
       return resource.process(a, b);
    }
}

2、策略

@Component
public class OptStrategyContext{
 

    private Map<String, Opt> strategyMap = new ConcurrentHashMap<>();
 
    @Autowired
    public OptStrategyContext(Map<String, TalkService> strategyMap) {
        this.strategyMap.clear();
        this.strategyMap.putAll(strategyMap);
    }
 
    public int apply(Sting opt, int a, int b) {
        return strategyMap.get(opt).apply(a, b);
    }
}

2.如何去除不必要的!=判空

• 空对象模式

public class MyParser implements Parser {
  private static Action NO_ACTION = new Action() {
    public void doSomething() { /* do nothing */ }
  };

  public Action findAction(String userInput) {
    // ...
    if ( /* we can't find any actions */ ) {
      return NO_ACTION;
    }
  }
}

然后便可以始终可以这么调用

ParserFactory.getParser().findAction(someInput).doSomething();

• Java8中使用Optional

Outer outer = new Outer();
if (outer != null && outer.nested != null && outer.nested.inner != null) {
    System.out.println(outer.nested.inner.foo);
}

我们可以通过利用 Java 8 的 Optional 类型来摆脱所有这些 null 检查。map 方法接收一个 Function 类型的 lambda 表达式，并自动将每个 function 的结果包装成一个 Optional 对象。这使我们能够在一行中进行多个 map 操作。Null 检查是在底层自动处理的。

Optional.of(new Outer())
    .map(Outer::getNested)
    .map(Nested::getInner)
    .map(Inner::getFoo)
    .ifPresent(System.out::println);

还有一种实现相同作用的方式就是通过利用一个 supplier 函数来解决嵌套路径的问题:

Outer obj = new Outer();
resolve(() -> obj.getNested().getInner().getFoo())
    .ifPresent(System.out::println);