Java中的基本加密算法不解密

我试图用Java实现TEA微型加密算法，对包含音频的字节[]进行编码，大小为512

下面是我的加密函数：

//Encrypt 64bit/8byte buffer with 128bit/16byte key
public byte[] encrypt(int[] data, int[] key) { 
    int x = data[0]; 
    int y = data[1]; 
    ByteBuffer encrypted = ByteBuffer.allocate(8);
    int sum = 0; 
    int constant = 0x9e3779b9; //magic constant

    for (int k = 0; k < 32; ++k) { 
        sum += constant; 
        x += (y << 4 & 0xfffffff0) + key[0] ^ y + sum ^ (y >> 5 & 0x7ffffff) + key[1]; 
        y += (x << 4 & 0xfffffff0) + key[2] ^ x + sum ^ (x >> 5 & 0x7ffffff) + key[3]; 
    }
    encrypted.putInt(x);
    encrypted.putInt(y);
    return encrypted.array();
}

并解密：

public byte[] decrypt(int[] data, int[] key) { 
    int x = data[0]; 
    int y = data[1]; 
    ByteBuffer decrypted = ByteBuffer.allocate(8);
    int sum = 0xC6EF3720; //32*delta
    int constant = 0x9e3779b9; //magic constant

    for (int k = 0; k < 32; ++k) { 
        x -= (x << 4 & 0xfffffff0) + key[2] ^ x + sum ^ (x >> 5 & 0x7ffffff) + key[3]; 
        y -= (y << 4 & 0xfffffff0) + key[0] ^ y + sum ^ (y >> 5 & 0x7ffffff) + key[1]; 
        sum -= constant; 
    }
    decrypted.putInt(x); 
    decrypted.putInt(y);
    return decrypted.array();
}

和我的加密呼叫：

ByteBuffer unwrapEncrypt = ByteBuffer.allocate(512);
int[] encryptionKey = {55555, 8888, 123857, 912029};

//block is a byte[] with length 512
ByteBuffer plainText = ByteBuffer.wrap(block);
for (int j = 0; j < block.length / 8; j++) {
    //Initiate array for int pairs
    int[] plainTextInts = new int[2];
    plainTextInts[0] = plainText.getInt();
    plainTextInts[1] = plainText.getInt();
    //Encrypt and store
    unwrapEncrypt.put(encrypt(plainTextInts, encryptionKey));
}

ByteBuffer audioToPlay = ByteBuffer.allocate(512);
int[] decryptionKey = {55555, 8888, 123857, 912029};

//audio is a byte[] with length 512
ByteBuffer cipherText = ByteBuffer.wrap(audio);
for (int j = 0; j < audio.length / 8; j++) {
    int[] plainTextInts = new int[2];
    //Initiate array for int pairs
    plainTextInts[0] = cipherText.getInt();
    plainTextInts[1] = cipherText.getInt();
    //Decrypt and store
    audioToPlay.put(decrypt(plainTextInts, decryptionKey));
}

和解密调用：

ByteBuffer unwrapEncrypt = ByteBuffer.allocate(512);
int[] encryptionKey = {55555, 8888, 123857, 912029};

//block is a byte[] with length 512
ByteBuffer plainText = ByteBuffer.wrap(block);
for (int j = 0; j < block.length / 8; j++) {
    //Initiate array for int pairs
    int[] plainTextInts = new int[2];
    plainTextInts[0] = plainText.getInt();
    plainTextInts[1] = plainText.getInt();
    //Encrypt and store
    unwrapEncrypt.put(encrypt(plainTextInts, encryptionKey));
}

ByteBuffer audioToPlay = ByteBuffer.allocate(512);
int[] decryptionKey = {55555, 8888, 123857, 912029};

//audio is a byte[] with length 512
ByteBuffer cipherText = ByteBuffer.wrap(audio);
for (int j = 0; j < audio.length / 8; j++) {
    int[] plainTextInts = new int[2];
    //Initiate array for int pairs
    plainTextInts[0] = cipherText.getInt();
    plainTextInts[1] = cipherText.getInt();
    //Decrypt and store
    audioToPlay.put(decrypt(plainTextInts, decryptionKey));
}

---

抱歉，代码太多了——我试过分析发送的音频和接收的解密数据——它们的长度都正确，只是完全不同。如果我删除这4块代码，音频是完美的。有人能认出发生了什么事吗？谢谢

我觉得您的decrpyt方法在与比较时有一个bug。必须在-=运算符的左侧交换x和y。以下几点似乎对我有用：

public byte[] decrypt(int[] data, int[] key) { 
    int x = data[0]; 
    int y = data[1]; 
    ByteBuffer decrypted = ByteBuffer.allocate(8);
    int sum = 0xC6EF3720; //32*delta
    int constant = 0x9e3779b9; //magic constant

    for (int k = 0; k < 32; ++k) { 
        y -= (x << 4 & 0xfffffff0) + key[2] ^ x + sum ^ (x >> 5 & 0x7ffffff) + key[3]; 
        x -= (y << 4 & 0xfffffff0) + key[0] ^ y + sum ^ (y >> 5 & 0x7ffffff) + key[1]; 
        sum -= constant; 
    }
    decrypted.putInt(x); 
    decrypted.putInt(y);
    return decrypted.array();
}

---

>>->>>>>？你是说在哪里？在右班。哦，我认为&优先。所以x>>5&0x7ffffff和y>>5&0x7ffffff->x>>>5&0x7ffffff和y>>5&0x7ffffff内部加密和解密？我认为x>>5&0x7ffffffff。请注意，我甚至还没有看过对茶的描述，但这对我来说似乎是不正确的。事实上，即使上面的说法似乎也不正确，因为对于这段代码，您不需要-。一步一步地测试。它应该是旋转的而不是移位的吗？嗯，我来看看。如果我能睁大眼睛，那就是。啊，我想这是另一个错误。请注意，解密工作并不意味着它是TEA的正确实现。您需要针对测试向量或已知良好的实现进行测试。而且，请注意，即使两个人在实现过程中也可能犯相同的错误。如果您有测试向量，那么很容易看到，如果它是设计师第一次实现，那么就不容易看到。