什么和& 0xff和MD5结构？ [英] What does & 0xff do And MD5 Structure?

问题描述

import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;

public class JavaMD5 {

    public static void main(String[] args) {
        String passwordToHash = "MyPassword123";
        String generatedPassword = null;
        try {
            MessageDigest md = MessageDigest.getInstance("MD5");
            md.update(passwordToHash.getBytes());
            byte[] bytes = md.digest();
            StringBuilder sb = new StringBuilder();
            for (int i = 0; i < bytes.length; i++) {
                sb.append(Integer.toString((bytes[i] & 0xff) + 0x100, 16).substring(1));
            }
            generatedPassword = sb.toString();

        } catch (NoSuchAlgorithmException e) {
            // TODO Auto-generated catch block
            e.printStackTrace();
        }
        System.out.println(generatedPassword);
    }
}

这一行是问题所在：

sb.append(Integer.toString((bytes[i] & 0xff) + 0x100, 16).substring(1));

这个结构中每个部分做了什么????

what does each part do in this structure????

谢谢，我很抱歉要求Beacuse我是java新手。

Thanks and I'm sorry for asking Beacuse I'm new in java.

推荐答案

大概是代码很清楚，这里唯一的谜是这个表达式：

Presumably most of the code is clear and the only mystery for you here is this expression:

(bytes[i] & 0xff) + 0x100

第一部分：

bytes[i] & 0xff

将位置 i 的字节扩大为一个 int 值，位位置为8-31，零。在Java中， byte 数据类型是有符号整数值，因此扩展符号扩展了该值。没有& 0xff ，大于0x7f的值最终将为负 int 值。其余的是相当明显的：它增加了0x100，它只是打开索引8处的位（因为它保证在中为0（bytes [i]& 0xff）。然后通过调用 Integer.toString（...，16） String 值>。

widens the byte at position i to an int value with zeros in bit positions 8-31. In Java, the byte data type is a signed integer value, so the widening sign-extends the value. Without the & 0xff, values greater than 0x7f would end up as negative int values. The rest is then fairly obvious: it adds 0x100, which simply turns on the bit at index 8 (since it is guaranteed to be 0 in (bytes[i] & 0xff). It is then converted to a hex String value by the call to Integer.toString(..., 16).

首先添加0x100然后剥离1的原因（由子串（1）完成调用，从子行到第1位开始的子字符串）是为了保证最终结果中有两个十六进制数字。否则，低于0x10的字节值在转换为十六进制时最终会成为一个字符的字符串。

The reason for first adding 0x100 and then stripping off the 1 (done by the substring(1) call, which takes the substring starting at position 1 through the end) is to guarantee two hex digits in the end result. Otherwise, byte values below 0x10 would end up as one-character strings when converted to hex.

这是否具有更好的表现（当然不是更清晰）是否值得商榷：

It's debatable whether all that has better performance (it certainly isn't clearer) than:

sb.append(String.format("%02x", bytes[i]));

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