首页 \ 问答 \ LinkedHashSet:hashCode()和equals()匹配,但contains()不匹配(LinkedHashSet: hashCode() and equals() match, but contains() doesn't)

LinkedHashSet:hashCode()和equals()匹配,但contains()不匹配(LinkedHashSet: hashCode() and equals() match, but contains() doesn't)

以下是如何可能的:

void contains(LinkedHashSet data, Object arg) {
    System.out.println(data.getClass()); // java.util.LinkedHashSet
    System.out.println(arg.hashCode() == data.iterator().next().hashCode()); // true
    System.out.println(arg.equals(data.iterator().next())); // true
    System.out.println(new ArrayList(data).contains(arg)); // true
    System.out.println(new HashSet(data).contains(arg)); // true
    System.out.println(new LinkedHashSet(data).contains(arg)); // true (!)
    System.out.println(data.contains(arg)); // false
}

难道我做错了什么?

显然,它并不总是发生(如果你创建一组琐碎的对象,你将无法重现它)。 但它总是发生在我的情况下更复杂的arg类。

编辑 :我没有在这里定义arg主要原因是它是相当大的类,Eclipse生成的hashCode跨越20行并且equals两倍。 而且我不认为它是相关的 - 只要它们对于这两个对象是相同的。


How is the following possible:

void contains(LinkedHashSet data, Object arg) {
    System.out.println(data.getClass()); // java.util.LinkedHashSet
    System.out.println(arg.hashCode() == data.iterator().next().hashCode()); // true
    System.out.println(arg.equals(data.iterator().next())); // true
    System.out.println(new ArrayList(data).contains(arg)); // true
    System.out.println(new HashSet(data).contains(arg)); // true
    System.out.println(new LinkedHashSet(data).contains(arg)); // true (!)
    System.out.println(data.contains(arg)); // false
}

Am I doing something wrong?

Obviously, it doesn't always happen (if you create a trivial set of Objects, you won't reproduce it). But it does always happen in my case with more complicated class of arg.

EDIT: The main reason why I don't define arg here is that's it's fairly big class, with Eclipse-generated hashCode that spans 20 lines and equals twice as long. And I don't think it's relevant - as long as they're equal for the two objects.


原文:https://stackoverflow.com/questions/8007723
更新时间:2023-12-07 06:12

最满意答案

React只会呈现JSX或字符串。 除非this.state.data是一个字符串,否则你需要将它转换为React渲染的东西。

您可以通过对数据进行字符串化来测试:

    <p className="App-intro">
      Este es el resultado de la consulta = <b>{JSON.stringify(this.state.data)}</b>
    </p>

如果this.state.data是一个数组,则需要将其映射到JSX元素。


React will only render JSX or strings. Unless this.state.data is a string, you will need to transform it into something React render.

You can test this by stringifying your data:

    <p className="App-intro">
      Este es el resultado de la consulta = <b>{JSON.stringify(this.state.data)}</b>
    </p>

If this.state.data is a array, you will need to map it to JSX elements.

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