本文介绍了是否真的缓存了java.lang.String的哈希码?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

String s1 = "String1";
System.out.println(s1.hashCode()); // return an integer i1


Field field = String.class.getDeclaredField("value");
field.setAccessible(true);
char[] value = (char[])field.get(s1);
value[0] = 'J';
value[1] = 'a';
value[2] = 'v';
value[3] = 'a';
value[4] = '1';
System.out.println(s1.hashCode()); // return same value of integer i1

这里甚至在我用反射的帮助改变了字符后,相同的哈希码值是主要的。

Here even after I changed the characters with the help of reflection, same hash code value is mainatained.

这里有什么我需要知道的吗?

Is there anything I need to know here?

推荐答案

字符串是不可变的。因此,没有必要重新计算哈希码。它在内部缓存在名为 hash 的字段中,类型为 int

A String is meant to be immutable. As such, there is no point having to recalculate the hashcode. It is cached internally in a field called hash of type int.

String#hashCode()实现为(Oracle JDK7)

String#hashCode() is implemented as (Oracle JDK7)

public int hashCode() {
    int h = hash;
    if (h == 0 && value.length > 0) {
        char val[] = value;

        for (int i = 0; i < value.length; i++) {
            h = 31 * h + val[i];
        }
        hash = h;
    }
    return h;
}

其中哈希最初的值为 0 。它只会在第一次调用方法时计算。

where hash initially has a value of 0. It will only be calculated the first time the method is called.

如评论中所述,使用反射会破坏对象的不变性。

As stated in the comments, using reflection breaks the immutability of the object.

这篇关于是否真的缓存了java.lang.String的哈希码?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-06 22:33