非空字符串的哈希码是否可以为零？ [英] Can a non-empty string have a hashcode of zero?

问题描述

通过非空，我的意思是在这个问题中至少包含一个非零字符的字符串。

作为参考，这里是实现：

pre $ 1493 public int hashCode（）{
1494 int h =散列;
1495 if（h == 0）{
1496 int off = offset;
1497 char val [] = value;
1498 int len = count;
1499
1500 for（int i = 0; i< len; i ++）{
1501 h = 31 * h + val [off ++];
1502}
1503 hash = h;
1504}
1505 return h;
1506}


并且在文档中指定了该算法。

在发生整数溢出之前，答案很简单：没有。但是我想知道的是，如果由于整数溢出，非空字符串可能具有零散列码？你可以构建一个吗？

我想要的是理论上的数学演示（或指向一个的链接）或构造算法。 $ b

解决方案

当然。例如，字符串 f5a5a608 的哈希码为零。

我发现通过简单的蛮力搜索：

  public static void main（String [] args）{
 long i = 0; 
 loop：while（true）{
 String s = Long.toHexString（i）; 
 if（s.hashCode（）== 0）{
 System.out.println（Found：'+ s +'）; 
 break循环; 
} 
 if（i％1000000 == 0）{
 System.out.println（checked：+ i）; 
} 
 i ++; 
 
 
 
 
 
 编辑：约瑟夫达西，他曾在JVM上工作过，甚至写过一个程序，可以用给定的哈希代码构建一个字符串（以测试在switch / case语句中实现字符串）通过反向运行哈希算法。
 
By "non-empty", I mean in this question a string which contains at least one non-zero character.

For reference, here's the hashCode implementation :
1493    public int hashCode() {
1494        int h = hash;
1495        if (h == 0) {
1496            int off = offset;
1497            char val[] = value;
1498            int len = count;
1499
1500            for (int i = 0; i < len; i++) {
1501                h = 31*h + val[off++];
1502            }
1503            hash = h;
1504        }
1505        return h;
1506    }
and the algorithm is specified in the documentation.

Before an integer overflow occurs, the answer is easy: it's no. But what I'd like to know is if, due to integer overflow, it's possible for a non-empty string to have a hashcode of zero? Can you construct one?

What I'm looking for would ideally be a mathematical demonstration (or a link to one) or a construction algorithm.
解决方案
Sure. The string f5a5a608 for example has a hashcode of zero.

I found that through a simple brute force search:
public static void main(String[] args){
    long i = 0;
    loop: while(true){
        String s = Long.toHexString(i);
        if(s.hashCode() == 0){
            System.out.println("Found: '"+s+"'");
            break loop;
        }
        if(i % 1000000==0){
            System.out.println("checked: "+i);              
        }
        i++;
    }       
}
Edit: Joseph Darcy, who worked on the JVM, even wrote a program that can construct a string with a given hashcode (to test the implementation of Strings in switch/case statements) by basically running the hash algorithm in reverse.

                        
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