合并 Java8 中的两个函数

Question

在isReadyToDeliver方法中，如果订单中的所有产品都可用（ProductState.AVAILABLE）并且订单状态已准备好发送（OrderState.READY_TO_SEND），方法必须returntrue。我写了两个部分，但我无法将它们组合成 return 短语，

我写了 return orderState.andThen(productState) 但收到此错误：

The method andThen(Function<? super Boolean,? extends V>) in the type Function<Order,Boolean> is not applicable for the arguments (Function<Order,Boolean>)

public class OrderFunctions  {

    public Function<Order, Boolean> isReadyToDeliver() {            
        Function<Order, Boolean> orderState = o -> o.getState() == OrderState.READY_TO_SEND;            
        Function<Order, Boolean>  productState = 
                o -> o.getProducts()
                    .stream()
                    .map(Product -> Product.getState())
                    .allMatch(Product -> Product == ProductState.AVAILABLE);

        return ????? ; 
       //return  orderState.andThen(productState);
       //error: The method andThen(Function<? super Boolean,? extends V>) in the type Function<Order,Boolean> is not applicable for the arguments (Function<Order,Boolean>)      
    }
}

如果还需要类：

enum OrderState {CONFIRMED, PAID, WAREHOUSE_PROCESSED, READY_TO_SEND, DELIVERED }

enum ProductType { NORMAL, BREAKABLE, PERISHABLE }

public class Order {

    private OrderState state;
    private List<Product> products = new ArrayList<>();

    public OrderState getState() {
        return state;
    }

    public void setState(OrderState state) {
        this.state = state;
    }

    public Order state(OrderState state) {
        this.state = state;
        return this;
    }

    public List<Product> getProducts() {
        return products;
    }

    public void setProducts(List<Product> products) {
        this.products = products;
    }

    public Order product(Product product) {
        if (products == null) {
            products = new ArrayList<>();
        }
        products.add(product);
        return this;
    }
}

public class Product {

    private String code;
    private String title;
    private ProductState state;

    public ProductState getState() {
        return state;
    }

    public void setState(ProductState state) {
        this.state = state;
    }

    public Product state(ProductState state) {
        this.state = state;
        return this;
    }
}

Answer 1

如果您将 isReadyToDeliver() 更改为 return Predicate<Order> 那么您将能够使用 .and(Predicate another) 函数组合两个谓词：

public Predicate<Order> isReadyToDeliver() {
    Predicate<Order> orderState = o -> o.getState() == OrderState.READY_TO_SEND;

    Predicate<Order> productState =
                o -> o.getProducts()
                   .stream()
                   .map(Product -> Product.getState())
                   .allMatch(Product -> Product == ProductState.AVAILABLE);

    return orderState.and(productState);
}

你的函数组合示例不起作用，因为当你组合函数 f 和 g 时，g 将 f 函数 return秒。在你的情况下它被打破了，因为 orderState 期望 Order 和 return Boolean 而在这种情况下 orderState.andThen() 期望一个函数将 Boolean 作为一个参数和 returns 其他东西。此要求未得到满足，因为 productState 预期 Order 并且 return 已 Boolean。这正是以下错误所说的：

error: The method andThen(Function) in the type Function is not applicable for the arguments (Function)

但是如果出于某种原因你想留在 Function<Order, Boolean> 那么你将有 return 像这样的 lambda：

public Function<Order, Boolean> isReadyToDeliver() {
    Function<Order, Boolean> orderState = o -> o.getState() == OrderState.READY_TO_SEND;

    Function<Order, Boolean> productState =
            o -> o.getProducts()
                    .stream()
                    .map(Product -> Product.getState())
                    .allMatch(Product -> Product == ProductState.AVAILABLE);


    return (order) -> orderState.apply(order) && productState.apply(order);
}

Answer 2

来自orderState和productState这两个函数您可以使用 &&（逻辑和）创建一个新的 lambda 表达式 return 它：

public Function<Order, Boolean> isReadyToDeliver() {

    Function<Order, Boolean> orderState = ...;

    Function<Order, Boolean>  productState = ...;


    return o -> orderState.apply(o) && productState.apply(o);
}

合并 Java8 中的两个函数

Combine two Functions in Java8

java

lambda

functional-programming

java-8

functional-interface