如何在Java中从字节[]计算互联网校验和

2022-09-03 00:26:55

我试图弄清楚如何计算Java中的Internet Checksum,它给我带来了无尽的痛苦。(我在位操作方面很糟糕。我在C#中找到了一个版本 计算一个互联网(又名IP,又名RFC791)在C#中校验和。但是,我尝试将其转换为Java并没有看到产生正确的结果。有人能看到我做错了什么吗?我怀疑数据类型有问题。

public long getValue() {
    byte[] buf = { (byte) 0xed, 0x2A, 0x44, 0x10, 0x03, 0x30};
    int length = buf.length;
    int i = 0;

    long sum = 0;
    long data = 0;
    while (length > 1) {
        data = 0;
        data = (((buf[i]) << 8) | ((buf[i + 1]) & 0xFF));

        sum += data;
        if ((sum & 0xFFFF0000) > 0) {
            sum = sum & 0xFFFF;
            sum += 1;
        }

        i += 2;
        length -= 2;
    }

    if (length > 0) {
        sum += (buf[i] << 8);
        // sum += buffer[i];
        if ((sum & 0xFFFF0000) > 0) {
            sum = sum & 0xFFFF;
            sum += 1;
        }
    }
    sum = ~sum;
    sum = sum & 0xFFFF;
    return sum;
}

答案 1

编辑以应用来自@Andy,@EJP,@RD等人的评论,并添加额外的测试用例以确保。

我使用了@Andys答案的组合(正确识别问题的位置),并更新了代码,以包括链接答案中提供的单元测试以及经过验证的消息校验和附加测试用例。

首先实现

package org.example.checksum;

public class InternetChecksum {

  /**
   * Calculate the Internet Checksum of a buffer (RFC 1071 - http://www.faqs.org/rfcs/rfc1071.html)
   * Algorithm is
   * 1) apply a 16-bit 1's complement sum over all octets (adjacent 8-bit pairs [A,B], final odd length is [A,0])
   * 2) apply 1's complement to this final sum
   *
   * Notes:
   * 1's complement is bitwise NOT of positive value.
   * Ensure that any carry bits are added back to avoid off-by-one errors
   *
   *
   * @param buf The message
   * @return The checksum
   */
  public long calculateChecksum(byte[] buf) {
    int length = buf.length;
    int i = 0;

    long sum = 0;
    long data;

    // Handle all pairs
    while (length > 1) {
      // Corrected to include @Andy's edits and various comments on Stack Overflow
      data = (((buf[i] << 8) & 0xFF00) | ((buf[i + 1]) & 0xFF));
      sum += data;
      // 1's complement carry bit correction in 16-bits (detecting sign extension)
      if ((sum & 0xFFFF0000) > 0) {
        sum = sum & 0xFFFF;
        sum += 1;
      }

      i += 2;
      length -= 2;
    }

    // Handle remaining byte in odd length buffers
    if (length > 0) {
      // Corrected to include @Andy's edits and various comments on Stack Overflow
      sum += (buf[i] << 8 & 0xFF00);
      // 1's complement carry bit correction in 16-bits (detecting sign extension)
      if ((sum & 0xFFFF0000) > 0) {
        sum = sum & 0xFFFF;
        sum += 1;
      }
    }

    // Final 1's complement value correction to 16-bits
    sum = ~sum;
    sum = sum & 0xFFFF;
    return sum;

  }

}

然后在 JUnit4 中进行单元测试

package org.example.checksum;

import org.junit.Test;

import static junit.framework.Assert.assertEquals;

public class InternetChecksumTest {
  @Test
  public void simplestValidValue() {
    InternetChecksum testObject = new InternetChecksum();

    byte[] buf = new byte[1]; // should work for any-length array of zeros
    long expected = 0xFFFF;

    long actual = testObject.calculateChecksum(buf);

    assertEquals(expected, actual);
  }

  @Test
  public void validSingleByteExtreme() {
    InternetChecksum testObject = new InternetChecksum();

    byte[] buf = new byte[]{(byte) 0xFF};
    long expected = 0xFF;

    long actual = testObject.calculateChecksum(buf);

    assertEquals(expected, actual);
  }

  @Test
  public void validMultiByteExtrema() {
    InternetChecksum testObject = new InternetChecksum();

    byte[] buf = new byte[]{0x00, (byte) 0xFF};
    long expected = 0xFF00;

    long actual = testObject.calculateChecksum(buf);

    assertEquals(expected, actual);
  }

  @Test
  public void validExampleMessage() {
    InternetChecksum testObject = new InternetChecksum();

    // Berkley example http://www.cs.berkeley.edu/~kfall/EE122/lec06/tsld023.htm
    // e3 4f 23 96 44 27 99 f3
    byte[] buf = {(byte) 0xe3, 0x4f, 0x23, (byte) 0x96, 0x44, 0x27, (byte) 0x99, (byte) 0xf3};

    long expected = 0x1aff;

    long actual = testObject.calculateChecksum(buf);

    assertEquals(expected, actual);
  }

  @Test
  public void validExampleEvenMessageWithCarryFromRFC1071() {
    InternetChecksum testObject = new InternetChecksum();

    // RFC1071 example http://www.ietf.org/rfc/rfc1071.txt
    // 00 01 f2 03 f4 f5 f6 f7
    byte[] buf = {(byte) 0x00, 0x01, (byte) 0xf2, (byte) 0x03, (byte) 0xf4, (byte) 0xf5, (byte) 0xf6, (byte) 0xf7};

    long expected = 0x220d;

    long actual = testObject.calculateChecksum(buf);

    assertEquals(expected, actual);

  }

}

答案 2

一个更短的版本如下:

long checksum(byte[] buf, int length) {
    int i = 0;
    long sum = 0;
    while (length > 0) {
        sum += (buf[i++]&0xff) << 8;
        if ((--length)==0) break;
        sum += (buf[i++]&0xff);
        --length;
    }

    return (~((sum & 0xFFFF)+(sum >> 16)))&0xFFFF;
}

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