使用 RSA Cipher 的线程安全

Thread safety with RSA Cipher

正如标题所说,我想知道如何在给定 RSA 算法的多线程环境中最好地使用 Cipher 实例。

我已经阅读了关于该主题的几个问题,并且从我收集到的内容中:

我的问题

代码示例:

public class RsaPrototype {

private static PrivateKey privKey;
private static Cipher cipher;
private static final String PRIVATE_KEY_PATH ="./privK.pem";

/*
 * ALGORITHM/BLOCKING_MODE/PADDING_SCHEMA
 */
private static final String CIPHER_SPECIFICATION = "RSA/None/NoPadding";
private static final String RSA_PROVIDER ="BC";

static {

    Security.addProvider(new BouncyCastleProvider());
    importPrivateKey();
    cipher = getCipher();

}

/**
 * Initializes cipher with RSA algorithm, without blocking mode and padding.
 * Implementation provider is bouncy castle.
 * 
 * @return cipher instance.
 */
private static Cipher getCipher() {

    try 
    {

            Cipher cipher = Cipher.getInstance(CIPHER_SPECIFICATION, RSA_PROVIDER);
            cipher.init(Cipher.DECRYPT_MODE, privKey);
            return cipher;

    } catch (NoSuchAlgorithmException | NoSuchPaddingException | NoSuchProviderException e) {

        throw new RuntimeException(e.getMessage());
    }
}

/**
 * Imports public key from the given .PEM file into application cache.
 */
private static void importPrivateKey() {

    try (BufferedReader reader = 
            new BufferedReader(new FileReader(PRIVATE_KEY_PATH));

        PEMParser pemParser = 
            new PEMParser(reader);) {


        privKey = new JcaPEMKeyConverter().getPrivateKey((PrivateKeyInfo) pemParser.readObject());
    }

    catch (IOException ignorable) {
         // not handled
        }
    }

public static String decrypt(byte[] encryptedText) {

    byte[] plainText;

    synchronized (cipher) {

         plainText = cipher.doFinal(encryptedText);
    }

    return new String(plainText, StandardCharsets.UTF_8);
  }
}

加密(和写入消息)本质上是同步的。使用多线程不是我的方法。考虑一条消息 aabb。使用多个线程可能会变成 abba 或 baba 或 abab 或 bbaa。请注意,密码的内部状态也是以这种方式同步的...要取出 aabb,您必须发送 aabb。

如果其他人读到这篇文章,则不需要为每次使用 RSA 的调用重新初始化密码。虽然,密码池可用于提高性能。

我已经编写了快速负载测试来验证这一点。

看来在cipher.doInit()上同步,用单个Cipher实例解密就够了

private static Queue<String> results = new ConcurrentLinkedQueue<String>();

@Test
public void test() throws InterruptedException {

    String plainText = "some text to verify data integrity";

    String encryptedText = Encrypt.encrypt(plainText);


    for (int i = 0; i < 5000; i++) {

        new Thread( () -> { results.add(Decrypt.decrypt(encryptedText)); })
        .start();;          
    }

    Thread.sleep(5000);

    assertTrue(results.size() == 5000);

    while(!results.isEmpty()) {

        assertTrue(results.poll().equals(plainText));
    }
}