gpt4 book ai didi

java - 设计时泛型的问题

转载 作者:行者123 更新时间:2023-12-01 09:25:07 25 4
gpt4 key购买 nike

我在设计过滤器框架时遇到了一些泛型问题。我收到编译错误。我不明白以及如何解决。我尝试过给予,但我认为这不是一个完美的方式。编译出错后

The method filter(String, Filter<Product>) in the type
CollectionMapper3<Product> is not applicable for the arguments
(String, Filter<Object>)

代码:

public interface Filter<T> {
public abstract boolean matches(T object);
}


public abstract class AttributeFilter3<T,C extends Comparable<C>> implements Filter<T> {
String property;
Comparable<C> comparable;

public AttributeFilter3(Comparable<C> comparable) {
this.comparable = comparable ;
}

public void setProperty(String property) {
this.property = property;
}

protected abstract boolean compare(int value);

@Override
public boolean matches(T t) {
return false;
}

public static <T,C extends Comparable<C>> Filter<T> EQUALS(Comparable<C> comparable) {
return new AttributeFilter4<T,C>(comparable) {
@Override
protected boolean compare(int result) {
return result == 0;
}
};
}

}


public class CollectionMapper3<T> {

private Collection<T> collection;

public CollectionMapper3(Collection<T> collection) {
this.collection = collection;

}

@SuppressWarnings("rawtypes")
public final CollectionMapper3<T> filter(String propertyName, Filter<T> filter) {
Collection<T> result = new ArrayList<>();
for (T t : collection) {
if (filter instanceof AttributeFilter3) {
((AttributeFilter3) filter).setProperty(propertyName);
if (filter.matches(t)) {
result.add(t);
}
}
}
collection = result;
return this;
}

public Collection<T> getResult() {
return collection;
}

}


public class Client3 {
public static void main(String[] args) {
Collection<Product> allProducts = MockUtil.getAllProducts();
// Here is the compilation error I am getting
Collection<Product> result = new CollectionMapper3<Product>(allProducts)
.filter("name", AttributeFilter3.EQUALS("sss")).filter("quantityInStock", AttributeFilter3.EQUALS(8))
.getResult();
System.out.println(result);

}
}






public class Product {

private String name;

public void setName(String name) {
this.name = name;
}
public String getName() {
return name;
}

@Override
public String toString() {
return name ;
}

}

最佳答案

您没有提供任何信息来推断T当您调用AttributeFilter3.EQUALS("sss")时,所以 Java 将使用 Object .

您可以提供类型参数:

AttributeFilter3.<String, String>EQUALS("sss")

或者您可以更新方法签名以接受类型 token :

public static <T,C extends Comparable<C>> Filter<T> EQUALS(Comparable<C> comparable, Class<T> ignored) {

然后

AttributeFilter3.EQUALS("sss", String.class)

关于java - 设计时泛型的问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/39900165/

25 4 0
Copyright 2021 - 2024 cfsdn All Rights Reserved 蜀ICP备2022000587号
广告合作:1813099741@qq.com 6ren.com