Question

所以我有一个学校项目，我遇到了一些麻烦。我们获得了加密方法，然后要求为它创建一个解密方法。

testDebugUnitTest

所以当字符串长度正好是16（2个块）时，我已经得到了解密方法。偶数块（8的倍数）的其他字符串不起作用，24,32 ......等。部分块也不起作用;如果我加密长度为18的字符串并再次解密，则最后2个字符总是错误的。任何帮助深表感谢！这是我的解密方法：

public static byte[] encrypt(byte[] plaintext, BlockCipher64 cipher, long IV) {
    if(plaintext.length <= 8)
        throw new IllegalArgumentException("plaintext must be longer than 8 bytes!");

    byte[] ciphertext = new byte[plaintext.length];
    int blocks = plaintext.length / 8;
    if(plaintext.length % 8 != 0) ++blocks;

    long prev = IV;
    for(int block = 0; block < blocks; ++block) {
        prev = cipher.encrypt(prev ^ longAt(plaintext, block * 8));
        storeLongAt(ciphertext, prev, block * 8);
    }

    // copy penultimate to last, then prev to penultimate (ciphertext stealing)
    int lastBlock = (blocks - 1) * 8;
    int secondLastBlock = (blocks - 2) * 8;
    storeLongAt(ciphertext, longAt(ciphertext, secondLastBlock), lastBlock);
    storeLongAt(ciphertext, prev, secondLastBlock);

    return ciphertext;
}

如果需要任何其他信息，请告诉我......我真的很难过这个。

编辑：其他方法和类 storeLongAt（）和longAt（）：

public static byte[] decrypt(byte[] ciphertext, BlockCipher64 cipher, long IV) {
    // code here
    // check for an illegal argument
    if(ciphertext.length <= 8)
        throw new IllegalArgumentException("ciphertext must be longer than 8 bytes!");

    // create a byte[] for the plaintext
    byte[] plaintext = new byte[ciphertext.length];

    // calculate how many blocks there are
    int blocks = plaintext.length / 8;
    if(plaintext.length % 8 != 0) ++blocks;

    // handle the last two blocks (which are special because of ciphertext stealing)
    int lastBlock = (blocks - 1) * 8;
    int secondLastBlock = (blocks - 2) * 8;
    long lBlock = longAt(ciphertext, lastBlock);
    int lBlockSize = ciphertext.length % 8;
    // if block sizes are even; swap, if block size is partial; 
    if (lBlockSize != 0) {
        //get blocks to switch
        byte[] NLB = new byte[lBlockSize];
        for (int i=secondLastBlock, j=0; i<secondLastBlock + lBlockSize; i++, j++) {
            NLB[j] = ciphertext[i];
        }
        byte[] NSLB = new byte[lBlockSize];
        for (int i=lastBlock, j=0; i<lastBlock + lBlockSize; i++, j++) {
            NSLB[j] = ciphertext[i];
        }
        //build ciphertext
        for (int i=secondLastBlock, j=0; i<secondLastBlock + lBlockSize; i++, j++) {
            ciphertext[i] = NSLB[j];
        }
        for (int i=lastBlock, j=0; i<lastBlock + lBlockSize; i++, j++) {
            ciphertext[i] = NLB[j];
        }
    } else {
        storeLongAt(ciphertext, longAt(ciphertext, secondLastBlock), lastBlock);
        storeLongAt(ciphertext, lBlock, secondLastBlock);
    }

    // loop over all other blocks, decrypting and xor'ing, and saving the results
    long prev = IV;
    for(int block = 0; block < blocks; ++block) {
        prev = cipher.decrypt(prev ^ longAt(ciphertext, block * 8));
        storeLongAt(plaintext, prev, block * 8);
    }

    return plaintext;
}

和BlockCipher64类：

public static void storeLongAt(byte[] b, long x, int pos) {
    byte[] c = BlockCipher64.longToBytes(x);
    for(int i = 0; i < 8 && (pos + i) < b.length; ++i) {
        b[pos + i] = c[i];
    }
}

public static long longAt(byte[] b, int pos) {
    return BlockCipher64.bytesToLong(Arrays.copyOfRange(b, pos, pos + 8));
}

Answer 1

您正在交换密文字节而不是明文字节。请注意，交换发生在之前加密最后一个块（及时）。如果您反向，则必须在解密最后一个块之后交换。

目前您正在创建无效的密文，并尝试解密该密文。

密文窃取Java

1 个答案: